Is there a possible geometric method to find length of this equilateral triangle?












11












$begingroup$



Problem Given that $AD parallel BC$, $|AB| = |AD|$, $angle A=120^{circ}$, $E$ is the midpoint of $AD$, point $F$ lies on $BD$, $triangle EFC$ is a equilateral triangle and $|AB|=4$, find the length $|EF|$.




Figure 1



Attempt At first glance, I thought it could be solved using a geometric method. I considered the law of sines/cosines, similar triangles, Pythagorean theorem, even Menelaus' theorem, however, got properties which contributed nothing to calculate $|EF|$.



What I've got after draw a line perpendicular to $BC$ through $E$





  • $triangle ABH$ and $triangle AHD$ are both equilateral triangles of length 4.

  • $triangle EFD sim triangle GEH$

  • $|EH|=2sqrt{3}$


Figure 2



Algebraic method Eventually, I've changed my mind to embrace algebra. I found it is easy to coordinate $E,A,B,D$ and $C$ is related to $F$ (rotation) and $B$ (same horizontal line). Make $E$ as the origin, $AD$ points to $x$-axis, $HE$ points to $y$-axis, we got




  • $E = (0,0)$

  • $A = (-2,0)$

  • $B = (-4,-2sqrt{3})$

  • $D = (2,0)$


Point $(x, y)$ in line $BD$ has $y=frac{1}{sqrt{3}}(x-2)$. Assume $F=(x_0,y_0)$, $C=(x_1,y_1)$, we can obtain $C$ by rotating $F$ around pivot $E$ $60^{circ}$ counter-clockwise



$$
begin{bmatrix}
x_1 \ y_1
end{bmatrix}
=
begin{bmatrix}
cos{theta} & -sin{theta} \
sin{theta} & cos{theta}
end{bmatrix}
begin{bmatrix}
x_0 \ y_0
end{bmatrix}
$$

, also we know that $BC$ is parallel to $x$-axis, then
$$
begin{align*}
y_1
& = sin{60^{circ}} x_0 + cos{60^{circ}} y_0 \
& = sin{60^{circ}} x_0 + cos{60^{circ}} frac{1}{sqrt{3}}(x_0-2) \
& = -2sqrt{3}
end{align*}
$$

, thus $F=(-frac{5}{2}, -frac{3sqrt{3}}{2})$, and finally $|EF|=sqrt{13}$



Thoughts afterwords I noticed that $F$ (through its coordinate) is actually the midpoint of $BK$. It may be a key point in geometric method, but I cannot prove it either.



Graph I made it in GeoGebra and it is shared. Please go and edit it to save your time if you have any idea.
Link: https://www.geogebra.org/graphing/yqhbzdem










share|cite|improve this question











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    11












    $begingroup$



    Problem Given that $AD parallel BC$, $|AB| = |AD|$, $angle A=120^{circ}$, $E$ is the midpoint of $AD$, point $F$ lies on $BD$, $triangle EFC$ is a equilateral triangle and $|AB|=4$, find the length $|EF|$.




    Figure 1



    Attempt At first glance, I thought it could be solved using a geometric method. I considered the law of sines/cosines, similar triangles, Pythagorean theorem, even Menelaus' theorem, however, got properties which contributed nothing to calculate $|EF|$.



    What I've got after draw a line perpendicular to $BC$ through $E$





    • $triangle ABH$ and $triangle AHD$ are both equilateral triangles of length 4.

    • $triangle EFD sim triangle GEH$

    • $|EH|=2sqrt{3}$


    Figure 2



    Algebraic method Eventually, I've changed my mind to embrace algebra. I found it is easy to coordinate $E,A,B,D$ and $C$ is related to $F$ (rotation) and $B$ (same horizontal line). Make $E$ as the origin, $AD$ points to $x$-axis, $HE$ points to $y$-axis, we got




    • $E = (0,0)$

    • $A = (-2,0)$

    • $B = (-4,-2sqrt{3})$

    • $D = (2,0)$


    Point $(x, y)$ in line $BD$ has $y=frac{1}{sqrt{3}}(x-2)$. Assume $F=(x_0,y_0)$, $C=(x_1,y_1)$, we can obtain $C$ by rotating $F$ around pivot $E$ $60^{circ}$ counter-clockwise



    $$
    begin{bmatrix}
    x_1 \ y_1
    end{bmatrix}
    =
    begin{bmatrix}
    cos{theta} & -sin{theta} \
    sin{theta} & cos{theta}
    end{bmatrix}
    begin{bmatrix}
    x_0 \ y_0
    end{bmatrix}
    $$

    , also we know that $BC$ is parallel to $x$-axis, then
    $$
    begin{align*}
    y_1
    & = sin{60^{circ}} x_0 + cos{60^{circ}} y_0 \
    & = sin{60^{circ}} x_0 + cos{60^{circ}} frac{1}{sqrt{3}}(x_0-2) \
    & = -2sqrt{3}
    end{align*}
    $$

    , thus $F=(-frac{5}{2}, -frac{3sqrt{3}}{2})$, and finally $|EF|=sqrt{13}$



    Thoughts afterwords I noticed that $F$ (through its coordinate) is actually the midpoint of $BK$. It may be a key point in geometric method, but I cannot prove it either.



    Graph I made it in GeoGebra and it is shared. Please go and edit it to save your time if you have any idea.
    Link: https://www.geogebra.org/graphing/yqhbzdem










    share|cite|improve this question











    $endgroup$















      11












      11








      11





      $begingroup$



      Problem Given that $AD parallel BC$, $|AB| = |AD|$, $angle A=120^{circ}$, $E$ is the midpoint of $AD$, point $F$ lies on $BD$, $triangle EFC$ is a equilateral triangle and $|AB|=4$, find the length $|EF|$.




      Figure 1



      Attempt At first glance, I thought it could be solved using a geometric method. I considered the law of sines/cosines, similar triangles, Pythagorean theorem, even Menelaus' theorem, however, got properties which contributed nothing to calculate $|EF|$.



      What I've got after draw a line perpendicular to $BC$ through $E$





      • $triangle ABH$ and $triangle AHD$ are both equilateral triangles of length 4.

      • $triangle EFD sim triangle GEH$

      • $|EH|=2sqrt{3}$


      Figure 2



      Algebraic method Eventually, I've changed my mind to embrace algebra. I found it is easy to coordinate $E,A,B,D$ and $C$ is related to $F$ (rotation) and $B$ (same horizontal line). Make $E$ as the origin, $AD$ points to $x$-axis, $HE$ points to $y$-axis, we got




      • $E = (0,0)$

      • $A = (-2,0)$

      • $B = (-4,-2sqrt{3})$

      • $D = (2,0)$


      Point $(x, y)$ in line $BD$ has $y=frac{1}{sqrt{3}}(x-2)$. Assume $F=(x_0,y_0)$, $C=(x_1,y_1)$, we can obtain $C$ by rotating $F$ around pivot $E$ $60^{circ}$ counter-clockwise



      $$
      begin{bmatrix}
      x_1 \ y_1
      end{bmatrix}
      =
      begin{bmatrix}
      cos{theta} & -sin{theta} \
      sin{theta} & cos{theta}
      end{bmatrix}
      begin{bmatrix}
      x_0 \ y_0
      end{bmatrix}
      $$

      , also we know that $BC$ is parallel to $x$-axis, then
      $$
      begin{align*}
      y_1
      & = sin{60^{circ}} x_0 + cos{60^{circ}} y_0 \
      & = sin{60^{circ}} x_0 + cos{60^{circ}} frac{1}{sqrt{3}}(x_0-2) \
      & = -2sqrt{3}
      end{align*}
      $$

      , thus $F=(-frac{5}{2}, -frac{3sqrt{3}}{2})$, and finally $|EF|=sqrt{13}$



      Thoughts afterwords I noticed that $F$ (through its coordinate) is actually the midpoint of $BK$. It may be a key point in geometric method, but I cannot prove it either.



      Graph I made it in GeoGebra and it is shared. Please go and edit it to save your time if you have any idea.
      Link: https://www.geogebra.org/graphing/yqhbzdem










      share|cite|improve this question











      $endgroup$





      Problem Given that $AD parallel BC$, $|AB| = |AD|$, $angle A=120^{circ}$, $E$ is the midpoint of $AD$, point $F$ lies on $BD$, $triangle EFC$ is a equilateral triangle and $|AB|=4$, find the length $|EF|$.




      Figure 1



      Attempt At first glance, I thought it could be solved using a geometric method. I considered the law of sines/cosines, similar triangles, Pythagorean theorem, even Menelaus' theorem, however, got properties which contributed nothing to calculate $|EF|$.



      What I've got after draw a line perpendicular to $BC$ through $E$





      • $triangle ABH$ and $triangle AHD$ are both equilateral triangles of length 4.

      • $triangle EFD sim triangle GEH$

      • $|EH|=2sqrt{3}$


      Figure 2



      Algebraic method Eventually, I've changed my mind to embrace algebra. I found it is easy to coordinate $E,A,B,D$ and $C$ is related to $F$ (rotation) and $B$ (same horizontal line). Make $E$ as the origin, $AD$ points to $x$-axis, $HE$ points to $y$-axis, we got




      • $E = (0,0)$

      • $A = (-2,0)$

      • $B = (-4,-2sqrt{3})$

      • $D = (2,0)$


      Point $(x, y)$ in line $BD$ has $y=frac{1}{sqrt{3}}(x-2)$. Assume $F=(x_0,y_0)$, $C=(x_1,y_1)$, we can obtain $C$ by rotating $F$ around pivot $E$ $60^{circ}$ counter-clockwise



      $$
      begin{bmatrix}
      x_1 \ y_1
      end{bmatrix}
      =
      begin{bmatrix}
      cos{theta} & -sin{theta} \
      sin{theta} & cos{theta}
      end{bmatrix}
      begin{bmatrix}
      x_0 \ y_0
      end{bmatrix}
      $$

      , also we know that $BC$ is parallel to $x$-axis, then
      $$
      begin{align*}
      y_1
      & = sin{60^{circ}} x_0 + cos{60^{circ}} y_0 \
      & = sin{60^{circ}} x_0 + cos{60^{circ}} frac{1}{sqrt{3}}(x_0-2) \
      & = -2sqrt{3}
      end{align*}
      $$

      , thus $F=(-frac{5}{2}, -frac{3sqrt{3}}{2})$, and finally $|EF|=sqrt{13}$



      Thoughts afterwords I noticed that $F$ (through its coordinate) is actually the midpoint of $BK$. It may be a key point in geometric method, but I cannot prove it either.



      Graph I made it in GeoGebra and it is shared. Please go and edit it to save your time if you have any idea.
      Link: https://www.geogebra.org/graphing/yqhbzdem







      geometry vectors euclidean-geometry analytic-geometry geometric-transformation






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      edited Jan 28 at 5:33







      alex4814

















      asked Jan 26 at 15:33









      alex4814alex4814

      1225




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          4 Answers
          4






          active

          oldest

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          14












          $begingroup$

          Since $$angle EDF = {1over 2}angle FCE $$ we see that $D$ is on a circle with center at $C$ and radius $CE =CF$ so $CD=CE$.



          enter image description here



          If $M$ is midpoint of $ED$ we have $$CE^2 = ME^2+CM^2 = 1+AG^2 = 13$$



          so $CE = sqrt{13}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Extremely beautiful! +1.
            $endgroup$
            – Michael Rozenberg
            Jan 27 at 5:27






          • 1




            $begingroup$
            I think the condition "$angle EDF = {1over 2}angle FCE$" is NOT strong enough to guarantee D lies on the same circle that is centered at C with radius = CF.
            $endgroup$
            – Mick
            Jan 28 at 3:54










          • $begingroup$
            At least explain why the first assertion is true. The explanation takes care to list the obvious radius equality, but doesn't even explain the logic of the circle.
            $endgroup$
            – C Perkins
            Jan 28 at 5:06










          • $begingroup$
            Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins
            $endgroup$
            – greedoid
            Jan 28 at 9:03








          • 1




            $begingroup$
            That clarifies one thing. Another requirement we need to say is both C and D should be on the same side of the line EF, but the original diagram clearly showed they are.
            $endgroup$
            – Mick
            Jan 28 at 17:28



















          9












          $begingroup$

          Let $P$ be the perpendicular foot of $E$ to $BD$. We find that $|EP|=sin(angle EDP )cdot|ED|=1$.
          enter image description here

          We also find that $triangle EPF$ is congruent to $triangle CHE $ implying that
          $$
          |EP|=|CH|=1.
          $$
          By Pythagorean theorem, it follows
          $$
          |EC|^2 =|EH|^2 +|CH|^2 =13,
          $$
          i.e. $$
          |EF|=|EC| =sqrt{13}.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you post a diagram? I cannot visually follow this at all.
            $endgroup$
            – The Great Duck
            Jan 26 at 17:47








          • 1




            $begingroup$
            @TheGreatDuck I've added a figure. I hope this will help.
            $endgroup$
            – Song
            Jan 26 at 18:19






          • 1




            $begingroup$
            This makes sense (+1), but it would help if the diagram didn't show quite so many unneeded points and lines. CD is useless-but-distracting even though mentioned in the problem, and the intersection points I, J, K, L, M, N, O are completely irrelevant.
            $endgroup$
            – Henning Makholm
            Jan 26 at 19:57










          • $begingroup$
            @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better ..
            $endgroup$
            – Song
            Jan 26 at 22:38










          • $begingroup$
            Much nicer this way.
            $endgroup$
            – Henning Makholm
            Jan 26 at 23:49



















          3












          $begingroup$

          I like the following way.



          Let $vec{AB}=vec{a}$, $vec{AD}=vec{b}$, $vec{BF}=pvec{BD}$ and $vec{BC}=kvec{AD}.$



          Thus, $$vec{FE}=-p(-vec{a}+vec{b})-vec{a}+frac{1}{2}vec{b}=(p-1)vec{a}+left(frac{1}{2}-pright)vec{b}$$ and
          $$vec{FC}=-p(-vec{a}+vec{b})+kvec{b}=pvec{a}+(k-p)vec{b}.$$



          Now, we obtain the following system:
          $$|vec{FE}|=|vec{FC}|$$ and
          $$frac{vec{FE}cdot vec{FC}}{|vec{FE}||vec{FC}|}=frac{1}{2}$$
          with variables $p$ and $k$.



          We can solve this system and the rest is smooth.






          share|cite|improve this answer









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            2












            $begingroup$

            Let $alpha=angle DEC$. We can apply the sine law to triangle $FED$:
            $$
            {EDoversin(90°-alpha)}={EFoversin30°}={FDoversin(alpha+60°)},
            $$

            that is:
            $$
            EF={1overcosalpha}quadtext{and}quad FD={2overcosalpha}sin(alpha+60°).
            $$

            Applying then the sine law to triangle $BFC$ one gets:
            $$
            FB={2overcosalpha}sin(alpha-60°)=4sqrt3-FD=4sqrt3-{2overcosalpha}sin(alpha+60°).
            $$

            From this it follows $tanalpha=2sqrt3$ and $EF^2=1/cos^2alpha=1+tan^2alpha=13$.






            share|cite|improve this answer









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              4 Answers
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              4 Answers
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              14












              $begingroup$

              Since $$angle EDF = {1over 2}angle FCE $$ we see that $D$ is on a circle with center at $C$ and radius $CE =CF$ so $CD=CE$.



              enter image description here



              If $M$ is midpoint of $ED$ we have $$CE^2 = ME^2+CM^2 = 1+AG^2 = 13$$



              so $CE = sqrt{13}$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Extremely beautiful! +1.
                $endgroup$
                – Michael Rozenberg
                Jan 27 at 5:27






              • 1




                $begingroup$
                I think the condition "$angle EDF = {1over 2}angle FCE$" is NOT strong enough to guarantee D lies on the same circle that is centered at C with radius = CF.
                $endgroup$
                – Mick
                Jan 28 at 3:54










              • $begingroup$
                At least explain why the first assertion is true. The explanation takes care to list the obvious radius equality, but doesn't even explain the logic of the circle.
                $endgroup$
                – C Perkins
                Jan 28 at 5:06










              • $begingroup$
                Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins
                $endgroup$
                – greedoid
                Jan 28 at 9:03








              • 1




                $begingroup$
                That clarifies one thing. Another requirement we need to say is both C and D should be on the same side of the line EF, but the original diagram clearly showed they are.
                $endgroup$
                – Mick
                Jan 28 at 17:28
















              14












              $begingroup$

              Since $$angle EDF = {1over 2}angle FCE $$ we see that $D$ is on a circle with center at $C$ and radius $CE =CF$ so $CD=CE$.



              enter image description here



              If $M$ is midpoint of $ED$ we have $$CE^2 = ME^2+CM^2 = 1+AG^2 = 13$$



              so $CE = sqrt{13}$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Extremely beautiful! +1.
                $endgroup$
                – Michael Rozenberg
                Jan 27 at 5:27






              • 1




                $begingroup$
                I think the condition "$angle EDF = {1over 2}angle FCE$" is NOT strong enough to guarantee D lies on the same circle that is centered at C with radius = CF.
                $endgroup$
                – Mick
                Jan 28 at 3:54










              • $begingroup$
                At least explain why the first assertion is true. The explanation takes care to list the obvious radius equality, but doesn't even explain the logic of the circle.
                $endgroup$
                – C Perkins
                Jan 28 at 5:06










              • $begingroup$
                Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins
                $endgroup$
                – greedoid
                Jan 28 at 9:03








              • 1




                $begingroup$
                That clarifies one thing. Another requirement we need to say is both C and D should be on the same side of the line EF, but the original diagram clearly showed they are.
                $endgroup$
                – Mick
                Jan 28 at 17:28














              14












              14








              14





              $begingroup$

              Since $$angle EDF = {1over 2}angle FCE $$ we see that $D$ is on a circle with center at $C$ and radius $CE =CF$ so $CD=CE$.



              enter image description here



              If $M$ is midpoint of $ED$ we have $$CE^2 = ME^2+CM^2 = 1+AG^2 = 13$$



              so $CE = sqrt{13}$.






              share|cite|improve this answer









              $endgroup$



              Since $$angle EDF = {1over 2}angle FCE $$ we see that $D$ is on a circle with center at $C$ and radius $CE =CF$ so $CD=CE$.



              enter image description here



              If $M$ is midpoint of $ED$ we have $$CE^2 = ME^2+CM^2 = 1+AG^2 = 13$$



              so $CE = sqrt{13}$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 26 at 20:41









              greedoidgreedoid

              45.5k1159114




              45.5k1159114












              • $begingroup$
                Extremely beautiful! +1.
                $endgroup$
                – Michael Rozenberg
                Jan 27 at 5:27






              • 1




                $begingroup$
                I think the condition "$angle EDF = {1over 2}angle FCE$" is NOT strong enough to guarantee D lies on the same circle that is centered at C with radius = CF.
                $endgroup$
                – Mick
                Jan 28 at 3:54










              • $begingroup$
                At least explain why the first assertion is true. The explanation takes care to list the obvious radius equality, but doesn't even explain the logic of the circle.
                $endgroup$
                – C Perkins
                Jan 28 at 5:06










              • $begingroup$
                Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins
                $endgroup$
                – greedoid
                Jan 28 at 9:03








              • 1




                $begingroup$
                That clarifies one thing. Another requirement we need to say is both C and D should be on the same side of the line EF, but the original diagram clearly showed they are.
                $endgroup$
                – Mick
                Jan 28 at 17:28


















              • $begingroup$
                Extremely beautiful! +1.
                $endgroup$
                – Michael Rozenberg
                Jan 27 at 5:27






              • 1




                $begingroup$
                I think the condition "$angle EDF = {1over 2}angle FCE$" is NOT strong enough to guarantee D lies on the same circle that is centered at C with radius = CF.
                $endgroup$
                – Mick
                Jan 28 at 3:54










              • $begingroup$
                At least explain why the first assertion is true. The explanation takes care to list the obvious radius equality, but doesn't even explain the logic of the circle.
                $endgroup$
                – C Perkins
                Jan 28 at 5:06










              • $begingroup$
                Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins
                $endgroup$
                – greedoid
                Jan 28 at 9:03








              • 1




                $begingroup$
                That clarifies one thing. Another requirement we need to say is both C and D should be on the same side of the line EF, but the original diagram clearly showed they are.
                $endgroup$
                – Mick
                Jan 28 at 17:28
















              $begingroup$
              Extremely beautiful! +1.
              $endgroup$
              – Michael Rozenberg
              Jan 27 at 5:27




              $begingroup$
              Extremely beautiful! +1.
              $endgroup$
              – Michael Rozenberg
              Jan 27 at 5:27




              1




              1




              $begingroup$
              I think the condition "$angle EDF = {1over 2}angle FCE$" is NOT strong enough to guarantee D lies on the same circle that is centered at C with radius = CF.
              $endgroup$
              – Mick
              Jan 28 at 3:54




              $begingroup$
              I think the condition "$angle EDF = {1over 2}angle FCE$" is NOT strong enough to guarantee D lies on the same circle that is centered at C with radius = CF.
              $endgroup$
              – Mick
              Jan 28 at 3:54












              $begingroup$
              At least explain why the first assertion is true. The explanation takes care to list the obvious radius equality, but doesn't even explain the logic of the circle.
              $endgroup$
              – C Perkins
              Jan 28 at 5:06




              $begingroup$
              At least explain why the first assertion is true. The explanation takes care to list the obvious radius equality, but doesn't even explain the logic of the circle.
              $endgroup$
              – C Perkins
              Jan 28 at 5:06












              $begingroup$
              Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins
              $endgroup$
              – greedoid
              Jan 28 at 9:03






              $begingroup$
              Yes, just that it is not. But if you read further you see $CF=CE$! @Mick @C Perkins
              $endgroup$
              – greedoid
              Jan 28 at 9:03






              1




              1




              $begingroup$
              That clarifies one thing. Another requirement we need to say is both C and D should be on the same side of the line EF, but the original diagram clearly showed they are.
              $endgroup$
              – Mick
              Jan 28 at 17:28




              $begingroup$
              That clarifies one thing. Another requirement we need to say is both C and D should be on the same side of the line EF, but the original diagram clearly showed they are.
              $endgroup$
              – Mick
              Jan 28 at 17:28











              9












              $begingroup$

              Let $P$ be the perpendicular foot of $E$ to $BD$. We find that $|EP|=sin(angle EDP )cdot|ED|=1$.
              enter image description here

              We also find that $triangle EPF$ is congruent to $triangle CHE $ implying that
              $$
              |EP|=|CH|=1.
              $$
              By Pythagorean theorem, it follows
              $$
              |EC|^2 =|EH|^2 +|CH|^2 =13,
              $$
              i.e. $$
              |EF|=|EC| =sqrt{13}.
              $$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Can you post a diagram? I cannot visually follow this at all.
                $endgroup$
                – The Great Duck
                Jan 26 at 17:47








              • 1




                $begingroup$
                @TheGreatDuck I've added a figure. I hope this will help.
                $endgroup$
                – Song
                Jan 26 at 18:19






              • 1




                $begingroup$
                This makes sense (+1), but it would help if the diagram didn't show quite so many unneeded points and lines. CD is useless-but-distracting even though mentioned in the problem, and the intersection points I, J, K, L, M, N, O are completely irrelevant.
                $endgroup$
                – Henning Makholm
                Jan 26 at 19:57










              • $begingroup$
                @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better ..
                $endgroup$
                – Song
                Jan 26 at 22:38










              • $begingroup$
                Much nicer this way.
                $endgroup$
                – Henning Makholm
                Jan 26 at 23:49
















              9












              $begingroup$

              Let $P$ be the perpendicular foot of $E$ to $BD$. We find that $|EP|=sin(angle EDP )cdot|ED|=1$.
              enter image description here

              We also find that $triangle EPF$ is congruent to $triangle CHE $ implying that
              $$
              |EP|=|CH|=1.
              $$
              By Pythagorean theorem, it follows
              $$
              |EC|^2 =|EH|^2 +|CH|^2 =13,
              $$
              i.e. $$
              |EF|=|EC| =sqrt{13}.
              $$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Can you post a diagram? I cannot visually follow this at all.
                $endgroup$
                – The Great Duck
                Jan 26 at 17:47








              • 1




                $begingroup$
                @TheGreatDuck I've added a figure. I hope this will help.
                $endgroup$
                – Song
                Jan 26 at 18:19






              • 1




                $begingroup$
                This makes sense (+1), but it would help if the diagram didn't show quite so many unneeded points and lines. CD is useless-but-distracting even though mentioned in the problem, and the intersection points I, J, K, L, M, N, O are completely irrelevant.
                $endgroup$
                – Henning Makholm
                Jan 26 at 19:57










              • $begingroup$
                @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better ..
                $endgroup$
                – Song
                Jan 26 at 22:38










              • $begingroup$
                Much nicer this way.
                $endgroup$
                – Henning Makholm
                Jan 26 at 23:49














              9












              9








              9





              $begingroup$

              Let $P$ be the perpendicular foot of $E$ to $BD$. We find that $|EP|=sin(angle EDP )cdot|ED|=1$.
              enter image description here

              We also find that $triangle EPF$ is congruent to $triangle CHE $ implying that
              $$
              |EP|=|CH|=1.
              $$
              By Pythagorean theorem, it follows
              $$
              |EC|^2 =|EH|^2 +|CH|^2 =13,
              $$
              i.e. $$
              |EF|=|EC| =sqrt{13}.
              $$






              share|cite|improve this answer











              $endgroup$



              Let $P$ be the perpendicular foot of $E$ to $BD$. We find that $|EP|=sin(angle EDP )cdot|ED|=1$.
              enter image description here

              We also find that $triangle EPF$ is congruent to $triangle CHE $ implying that
              $$
              |EP|=|CH|=1.
              $$
              By Pythagorean theorem, it follows
              $$
              |EC|^2 =|EH|^2 +|CH|^2 =13,
              $$
              i.e. $$
              |EF|=|EC| =sqrt{13}.
              $$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 26 at 22:23

























              answered Jan 26 at 16:15









              SongSong

              16.4k1741




              16.4k1741












              • $begingroup$
                Can you post a diagram? I cannot visually follow this at all.
                $endgroup$
                – The Great Duck
                Jan 26 at 17:47








              • 1




                $begingroup$
                @TheGreatDuck I've added a figure. I hope this will help.
                $endgroup$
                – Song
                Jan 26 at 18:19






              • 1




                $begingroup$
                This makes sense (+1), but it would help if the diagram didn't show quite so many unneeded points and lines. CD is useless-but-distracting even though mentioned in the problem, and the intersection points I, J, K, L, M, N, O are completely irrelevant.
                $endgroup$
                – Henning Makholm
                Jan 26 at 19:57










              • $begingroup$
                @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better ..
                $endgroup$
                – Song
                Jan 26 at 22:38










              • $begingroup$
                Much nicer this way.
                $endgroup$
                – Henning Makholm
                Jan 26 at 23:49


















              • $begingroup$
                Can you post a diagram? I cannot visually follow this at all.
                $endgroup$
                – The Great Duck
                Jan 26 at 17:47








              • 1




                $begingroup$
                @TheGreatDuck I've added a figure. I hope this will help.
                $endgroup$
                – Song
                Jan 26 at 18:19






              • 1




                $begingroup$
                This makes sense (+1), but it would help if the diagram didn't show quite so many unneeded points and lines. CD is useless-but-distracting even though mentioned in the problem, and the intersection points I, J, K, L, M, N, O are completely irrelevant.
                $endgroup$
                – Henning Makholm
                Jan 26 at 19:57










              • $begingroup$
                @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better ..
                $endgroup$
                – Song
                Jan 26 at 22:38










              • $begingroup$
                Much nicer this way.
                $endgroup$
                – Henning Makholm
                Jan 26 at 23:49
















              $begingroup$
              Can you post a diagram? I cannot visually follow this at all.
              $endgroup$
              – The Great Duck
              Jan 26 at 17:47






              $begingroup$
              Can you post a diagram? I cannot visually follow this at all.
              $endgroup$
              – The Great Duck
              Jan 26 at 17:47






              1




              1




              $begingroup$
              @TheGreatDuck I've added a figure. I hope this will help.
              $endgroup$
              – Song
              Jan 26 at 18:19




              $begingroup$
              @TheGreatDuck I've added a figure. I hope this will help.
              $endgroup$
              – Song
              Jan 26 at 18:19




              1




              1




              $begingroup$
              This makes sense (+1), but it would help if the diagram didn't show quite so many unneeded points and lines. CD is useless-but-distracting even though mentioned in the problem, and the intersection points I, J, K, L, M, N, O are completely irrelevant.
              $endgroup$
              – Henning Makholm
              Jan 26 at 19:57




              $begingroup$
              This makes sense (+1), but it would help if the diagram didn't show quite so many unneeded points and lines. CD is useless-but-distracting even though mentioned in the problem, and the intersection points I, J, K, L, M, N, O are completely irrelevant.
              $endgroup$
              – Henning Makholm
              Jan 26 at 19:57












              $begingroup$
              @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better ..
              $endgroup$
              – Song
              Jan 26 at 22:38




              $begingroup$
              @HenningMakholm Thank you for suggesting, sir. I've updated the figure. I hope this is better ..
              $endgroup$
              – Song
              Jan 26 at 22:38












              $begingroup$
              Much nicer this way.
              $endgroup$
              – Henning Makholm
              Jan 26 at 23:49




              $begingroup$
              Much nicer this way.
              $endgroup$
              – Henning Makholm
              Jan 26 at 23:49











              3












              $begingroup$

              I like the following way.



              Let $vec{AB}=vec{a}$, $vec{AD}=vec{b}$, $vec{BF}=pvec{BD}$ and $vec{BC}=kvec{AD}.$



              Thus, $$vec{FE}=-p(-vec{a}+vec{b})-vec{a}+frac{1}{2}vec{b}=(p-1)vec{a}+left(frac{1}{2}-pright)vec{b}$$ and
              $$vec{FC}=-p(-vec{a}+vec{b})+kvec{b}=pvec{a}+(k-p)vec{b}.$$



              Now, we obtain the following system:
              $$|vec{FE}|=|vec{FC}|$$ and
              $$frac{vec{FE}cdot vec{FC}}{|vec{FE}||vec{FC}|}=frac{1}{2}$$
              with variables $p$ and $k$.



              We can solve this system and the rest is smooth.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                I like the following way.



                Let $vec{AB}=vec{a}$, $vec{AD}=vec{b}$, $vec{BF}=pvec{BD}$ and $vec{BC}=kvec{AD}.$



                Thus, $$vec{FE}=-p(-vec{a}+vec{b})-vec{a}+frac{1}{2}vec{b}=(p-1)vec{a}+left(frac{1}{2}-pright)vec{b}$$ and
                $$vec{FC}=-p(-vec{a}+vec{b})+kvec{b}=pvec{a}+(k-p)vec{b}.$$



                Now, we obtain the following system:
                $$|vec{FE}|=|vec{FC}|$$ and
                $$frac{vec{FE}cdot vec{FC}}{|vec{FE}||vec{FC}|}=frac{1}{2}$$
                with variables $p$ and $k$.



                We can solve this system and the rest is smooth.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  I like the following way.



                  Let $vec{AB}=vec{a}$, $vec{AD}=vec{b}$, $vec{BF}=pvec{BD}$ and $vec{BC}=kvec{AD}.$



                  Thus, $$vec{FE}=-p(-vec{a}+vec{b})-vec{a}+frac{1}{2}vec{b}=(p-1)vec{a}+left(frac{1}{2}-pright)vec{b}$$ and
                  $$vec{FC}=-p(-vec{a}+vec{b})+kvec{b}=pvec{a}+(k-p)vec{b}.$$



                  Now, we obtain the following system:
                  $$|vec{FE}|=|vec{FC}|$$ and
                  $$frac{vec{FE}cdot vec{FC}}{|vec{FE}||vec{FC}|}=frac{1}{2}$$
                  with variables $p$ and $k$.



                  We can solve this system and the rest is smooth.






                  share|cite|improve this answer









                  $endgroup$



                  I like the following way.



                  Let $vec{AB}=vec{a}$, $vec{AD}=vec{b}$, $vec{BF}=pvec{BD}$ and $vec{BC}=kvec{AD}.$



                  Thus, $$vec{FE}=-p(-vec{a}+vec{b})-vec{a}+frac{1}{2}vec{b}=(p-1)vec{a}+left(frac{1}{2}-pright)vec{b}$$ and
                  $$vec{FC}=-p(-vec{a}+vec{b})+kvec{b}=pvec{a}+(k-p)vec{b}.$$



                  Now, we obtain the following system:
                  $$|vec{FE}|=|vec{FC}|$$ and
                  $$frac{vec{FE}cdot vec{FC}}{|vec{FE}||vec{FC}|}=frac{1}{2}$$
                  with variables $p$ and $k$.



                  We can solve this system and the rest is smooth.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 26 at 15:55









                  Michael RozenbergMichael Rozenberg

                  106k1893198




                  106k1893198























                      2












                      $begingroup$

                      Let $alpha=angle DEC$. We can apply the sine law to triangle $FED$:
                      $$
                      {EDoversin(90°-alpha)}={EFoversin30°}={FDoversin(alpha+60°)},
                      $$

                      that is:
                      $$
                      EF={1overcosalpha}quadtext{and}quad FD={2overcosalpha}sin(alpha+60°).
                      $$

                      Applying then the sine law to triangle $BFC$ one gets:
                      $$
                      FB={2overcosalpha}sin(alpha-60°)=4sqrt3-FD=4sqrt3-{2overcosalpha}sin(alpha+60°).
                      $$

                      From this it follows $tanalpha=2sqrt3$ and $EF^2=1/cos^2alpha=1+tan^2alpha=13$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Let $alpha=angle DEC$. We can apply the sine law to triangle $FED$:
                        $$
                        {EDoversin(90°-alpha)}={EFoversin30°}={FDoversin(alpha+60°)},
                        $$

                        that is:
                        $$
                        EF={1overcosalpha}quadtext{and}quad FD={2overcosalpha}sin(alpha+60°).
                        $$

                        Applying then the sine law to triangle $BFC$ one gets:
                        $$
                        FB={2overcosalpha}sin(alpha-60°)=4sqrt3-FD=4sqrt3-{2overcosalpha}sin(alpha+60°).
                        $$

                        From this it follows $tanalpha=2sqrt3$ and $EF^2=1/cos^2alpha=1+tan^2alpha=13$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Let $alpha=angle DEC$. We can apply the sine law to triangle $FED$:
                          $$
                          {EDoversin(90°-alpha)}={EFoversin30°}={FDoversin(alpha+60°)},
                          $$

                          that is:
                          $$
                          EF={1overcosalpha}quadtext{and}quad FD={2overcosalpha}sin(alpha+60°).
                          $$

                          Applying then the sine law to triangle $BFC$ one gets:
                          $$
                          FB={2overcosalpha}sin(alpha-60°)=4sqrt3-FD=4sqrt3-{2overcosalpha}sin(alpha+60°).
                          $$

                          From this it follows $tanalpha=2sqrt3$ and $EF^2=1/cos^2alpha=1+tan^2alpha=13$.






                          share|cite|improve this answer









                          $endgroup$



                          Let $alpha=angle DEC$. We can apply the sine law to triangle $FED$:
                          $$
                          {EDoversin(90°-alpha)}={EFoversin30°}={FDoversin(alpha+60°)},
                          $$

                          that is:
                          $$
                          EF={1overcosalpha}quadtext{and}quad FD={2overcosalpha}sin(alpha+60°).
                          $$

                          Applying then the sine law to triangle $BFC$ one gets:
                          $$
                          FB={2overcosalpha}sin(alpha-60°)=4sqrt3-FD=4sqrt3-{2overcosalpha}sin(alpha+60°).
                          $$

                          From this it follows $tanalpha=2sqrt3$ and $EF^2=1/cos^2alpha=1+tan^2alpha=13$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 26 at 16:30









                          AretinoAretino

                          24.7k21444




                          24.7k21444






























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