Prove that triangle $XYZ$ is equilateral












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$begingroup$


Let $ABC$ be an acute angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $ACB$ and $ABC$ with the line $DE$ and let $Z$ be the midpoint of the side $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $angle A = 60^o$.



I dont know why, but it seems to me that $Delta ADE$ and $Delta XYZ$ are similar (or maybe congruent : ). Is it true? Or no? Please help.










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  • $begingroup$
    A little playing around with Geogebra seems to confirm your intuition about similar triangles. Perhaps someone can give a hint toward a proof.
    $endgroup$
    – David K
    Jan 12 at 17:51
















1












$begingroup$


Let $ABC$ be an acute angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $ACB$ and $ABC$ with the line $DE$ and let $Z$ be the midpoint of the side $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $angle A = 60^o$.



I dont know why, but it seems to me that $Delta ADE$ and $Delta XYZ$ are similar (or maybe congruent : ). Is it true? Or no? Please help.










share|cite|improve this question











$endgroup$












  • $begingroup$
    A little playing around with Geogebra seems to confirm your intuition about similar triangles. Perhaps someone can give a hint toward a proof.
    $endgroup$
    – David K
    Jan 12 at 17:51














1












1








1


2



$begingroup$


Let $ABC$ be an acute angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $ACB$ and $ABC$ with the line $DE$ and let $Z$ be the midpoint of the side $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $angle A = 60^o$.



I dont know why, but it seems to me that $Delta ADE$ and $Delta XYZ$ are similar (or maybe congruent : ). Is it true? Or no? Please help.










share|cite|improve this question











$endgroup$




Let $ABC$ be an acute angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $ACB$ and $ABC$ with the line $DE$ and let $Z$ be the midpoint of the side $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $angle A = 60^o$.



I dont know why, but it seems to me that $Delta ADE$ and $Delta XYZ$ are similar (or maybe congruent : ). Is it true? Or no? Please help.







geometry






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share|cite|improve this question













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edited Jan 12 at 16:33







Yellow

















asked Jan 12 at 16:23









YellowYellow

16011




16011












  • $begingroup$
    A little playing around with Geogebra seems to confirm your intuition about similar triangles. Perhaps someone can give a hint toward a proof.
    $endgroup$
    – David K
    Jan 12 at 17:51


















  • $begingroup$
    A little playing around with Geogebra seems to confirm your intuition about similar triangles. Perhaps someone can give a hint toward a proof.
    $endgroup$
    – David K
    Jan 12 at 17:51
















$begingroup$
A little playing around with Geogebra seems to confirm your intuition about similar triangles. Perhaps someone can give a hint toward a proof.
$endgroup$
– David K
Jan 12 at 17:51




$begingroup$
A little playing around with Geogebra seems to confirm your intuition about similar triangles. Perhaps someone can give a hint toward a proof.
$endgroup$
– David K
Jan 12 at 17:51










2 Answers
2






active

oldest

votes


















3












$begingroup$

enter image description here



Let us first show that $angle BXC=angle BYC=90^circ$.



Notice that triangle $ADE$ is isosceles so $angle AED=90^circ-alpha/2$. It means that $angle DEC=angle XEC=90^circ+alpha/2$. We also know that $angle ECX=gamma/2$. From triangle $XEC$:



$$angle CXE=180^circ-angle XEC-angle ECX=180^circ-(90^circ+alpha/2)-gamma/2=beta/2$$



It follows immediatelly that $angle DXI=180-beta/2$ and $angle DXI+angle DBI=180^circ$. And therefore, quadrialteral $BIXD$ is concyclic. Because of that:



$$angle BXC=angle BXI=angle BDI=90^circtag{1}$$



In a similar way we can show that:



$$angle BYC=90^circtag{2}$$



Because of (1) and (2) points $X$ and $Y$ must be on a circle with diameter BC with center $Z$. So triangle $XYZ$ is isosceles with $ZX=ZY$.



Now:



$$angle XZY=2angle XBY=2(angle XBC-angle IBC)=2(90^circ-gamma/2-beta/2)=alpha$$



So triangle $XYZ$ is equilateral if and only if $alpha=60^circ$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Let
    $$A=(0, a)\
    B=(-b, 0)\
    C=(b, 0)$$

    and thus
    $$Z=(0, 0)$$
    then we get
    $$tan(angle ABC)=frac ab$$
    and thus
    $$tan(frac 12 angle ABC)=frac{a/b}{1+sqrt{1+a^2/b^2}}=frac a{b+sqrt{a^2+b^2}}$$



    So we can deduce the center $M$ of the incircle to be
    $$M=(0, frac {ab}{b+sqrt{a^2+b^2}})$$



    Now define lines $g$ and $h$ by
    $$g=overline{AC}: y=-frac ab x+a\
    h=overline{ME}: y=frac ba x+frac {ab}{b+sqrt{a^2+b^2}}$$

    Equating those will then provide
    $$frac {a^2+b^2}{ab}cdot x=frac {ab+asqrt{a^2+b^2}-ab}{b+sqrt{a^2+b^2}}$$
    or
    $$x=frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}$$
    Inserting $x$ into $h$ further provides
    $$y=frac{ab^2}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}+frac{ab}{b+sqrt{a^2+b^2}}cdotfrac{sqrt{a^2+b^2}}{sqrt{a^2+b^2}}=frac{ab}{sqrt{a^2+b^2}}$$
    Thus we have
    $$E=(frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})$$



    Now define line $k$ to be
    $$k=overline{BM}: y=frac a{b+sqrt{a^2+b^2}} (x+b)$$
    and intersecting that with $overline{DE}$, i.e. equating it with the $y$ value of $E$, provides
    $$frac a{b+sqrt{a^2+b^2}} (x+b)=frac{ab}{sqrt{a^2+b^2}}\
    x+b=frac{b(b+sqrt{a^2+b^2})}{sqrt{a^2+b^2}}=frac{b^2}{sqrt{a^2+b^2}}+b\
    x=frac{b^2}{sqrt{a^2+b^2}}$$



    Thus we have calculated $Y$ to be
    $$Y=(frac{b^2}{sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})=frac b{sqrt{a^2+b^2}} (b, a)$$
    and this finally proves your conjecture:
    $$overline{AB}paralleloverline{ZY}$$
    q.e. $ABC$ and $XYZ$ are indeed similar triangles, provided $ABC$ was an isoceles triangle, as asumed by the chosen coordinatisation.



    --- rk






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for the solution, but I like pure geometric proofs (I rarely prefer trig over pure geometry, but at times I’m forced to use trig) more than any other...
      $endgroup$
      – Yellow
      Jan 16 at 17:49











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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

    votes









    3












    $begingroup$

    enter image description here



    Let us first show that $angle BXC=angle BYC=90^circ$.



    Notice that triangle $ADE$ is isosceles so $angle AED=90^circ-alpha/2$. It means that $angle DEC=angle XEC=90^circ+alpha/2$. We also know that $angle ECX=gamma/2$. From triangle $XEC$:



    $$angle CXE=180^circ-angle XEC-angle ECX=180^circ-(90^circ+alpha/2)-gamma/2=beta/2$$



    It follows immediatelly that $angle DXI=180-beta/2$ and $angle DXI+angle DBI=180^circ$. And therefore, quadrialteral $BIXD$ is concyclic. Because of that:



    $$angle BXC=angle BXI=angle BDI=90^circtag{1}$$



    In a similar way we can show that:



    $$angle BYC=90^circtag{2}$$



    Because of (1) and (2) points $X$ and $Y$ must be on a circle with diameter BC with center $Z$. So triangle $XYZ$ is isosceles with $ZX=ZY$.



    Now:



    $$angle XZY=2angle XBY=2(angle XBC-angle IBC)=2(90^circ-gamma/2-beta/2)=alpha$$



    So triangle $XYZ$ is equilateral if and only if $alpha=60^circ$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      enter image description here



      Let us first show that $angle BXC=angle BYC=90^circ$.



      Notice that triangle $ADE$ is isosceles so $angle AED=90^circ-alpha/2$. It means that $angle DEC=angle XEC=90^circ+alpha/2$. We also know that $angle ECX=gamma/2$. From triangle $XEC$:



      $$angle CXE=180^circ-angle XEC-angle ECX=180^circ-(90^circ+alpha/2)-gamma/2=beta/2$$



      It follows immediatelly that $angle DXI=180-beta/2$ and $angle DXI+angle DBI=180^circ$. And therefore, quadrialteral $BIXD$ is concyclic. Because of that:



      $$angle BXC=angle BXI=angle BDI=90^circtag{1}$$



      In a similar way we can show that:



      $$angle BYC=90^circtag{2}$$



      Because of (1) and (2) points $X$ and $Y$ must be on a circle with diameter BC with center $Z$. So triangle $XYZ$ is isosceles with $ZX=ZY$.



      Now:



      $$angle XZY=2angle XBY=2(angle XBC-angle IBC)=2(90^circ-gamma/2-beta/2)=alpha$$



      So triangle $XYZ$ is equilateral if and only if $alpha=60^circ$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        enter image description here



        Let us first show that $angle BXC=angle BYC=90^circ$.



        Notice that triangle $ADE$ is isosceles so $angle AED=90^circ-alpha/2$. It means that $angle DEC=angle XEC=90^circ+alpha/2$. We also know that $angle ECX=gamma/2$. From triangle $XEC$:



        $$angle CXE=180^circ-angle XEC-angle ECX=180^circ-(90^circ+alpha/2)-gamma/2=beta/2$$



        It follows immediatelly that $angle DXI=180-beta/2$ and $angle DXI+angle DBI=180^circ$. And therefore, quadrialteral $BIXD$ is concyclic. Because of that:



        $$angle BXC=angle BXI=angle BDI=90^circtag{1}$$



        In a similar way we can show that:



        $$angle BYC=90^circtag{2}$$



        Because of (1) and (2) points $X$ and $Y$ must be on a circle with diameter BC with center $Z$. So triangle $XYZ$ is isosceles with $ZX=ZY$.



        Now:



        $$angle XZY=2angle XBY=2(angle XBC-angle IBC)=2(90^circ-gamma/2-beta/2)=alpha$$



        So triangle $XYZ$ is equilateral if and only if $alpha=60^circ$.






        share|cite|improve this answer











        $endgroup$



        enter image description here



        Let us first show that $angle BXC=angle BYC=90^circ$.



        Notice that triangle $ADE$ is isosceles so $angle AED=90^circ-alpha/2$. It means that $angle DEC=angle XEC=90^circ+alpha/2$. We also know that $angle ECX=gamma/2$. From triangle $XEC$:



        $$angle CXE=180^circ-angle XEC-angle ECX=180^circ-(90^circ+alpha/2)-gamma/2=beta/2$$



        It follows immediatelly that $angle DXI=180-beta/2$ and $angle DXI+angle DBI=180^circ$. And therefore, quadrialteral $BIXD$ is concyclic. Because of that:



        $$angle BXC=angle BXI=angle BDI=90^circtag{1}$$



        In a similar way we can show that:



        $$angle BYC=90^circtag{2}$$



        Because of (1) and (2) points $X$ and $Y$ must be on a circle with diameter BC with center $Z$. So triangle $XYZ$ is isosceles with $ZX=ZY$.



        Now:



        $$angle XZY=2angle XBY=2(angle XBC-angle IBC)=2(90^circ-gamma/2-beta/2)=alpha$$



        So triangle $XYZ$ is equilateral if and only if $alpha=60^circ$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 19:37

























        answered Jan 12 at 19:32









        OldboyOldboy

        8,67911036




        8,67911036























            1












            $begingroup$

            Let
            $$A=(0, a)\
            B=(-b, 0)\
            C=(b, 0)$$

            and thus
            $$Z=(0, 0)$$
            then we get
            $$tan(angle ABC)=frac ab$$
            and thus
            $$tan(frac 12 angle ABC)=frac{a/b}{1+sqrt{1+a^2/b^2}}=frac a{b+sqrt{a^2+b^2}}$$



            So we can deduce the center $M$ of the incircle to be
            $$M=(0, frac {ab}{b+sqrt{a^2+b^2}})$$



            Now define lines $g$ and $h$ by
            $$g=overline{AC}: y=-frac ab x+a\
            h=overline{ME}: y=frac ba x+frac {ab}{b+sqrt{a^2+b^2}}$$

            Equating those will then provide
            $$frac {a^2+b^2}{ab}cdot x=frac {ab+asqrt{a^2+b^2}-ab}{b+sqrt{a^2+b^2}}$$
            or
            $$x=frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}$$
            Inserting $x$ into $h$ further provides
            $$y=frac{ab^2}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}+frac{ab}{b+sqrt{a^2+b^2}}cdotfrac{sqrt{a^2+b^2}}{sqrt{a^2+b^2}}=frac{ab}{sqrt{a^2+b^2}}$$
            Thus we have
            $$E=(frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})$$



            Now define line $k$ to be
            $$k=overline{BM}: y=frac a{b+sqrt{a^2+b^2}} (x+b)$$
            and intersecting that with $overline{DE}$, i.e. equating it with the $y$ value of $E$, provides
            $$frac a{b+sqrt{a^2+b^2}} (x+b)=frac{ab}{sqrt{a^2+b^2}}\
            x+b=frac{b(b+sqrt{a^2+b^2})}{sqrt{a^2+b^2}}=frac{b^2}{sqrt{a^2+b^2}}+b\
            x=frac{b^2}{sqrt{a^2+b^2}}$$



            Thus we have calculated $Y$ to be
            $$Y=(frac{b^2}{sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})=frac b{sqrt{a^2+b^2}} (b, a)$$
            and this finally proves your conjecture:
            $$overline{AB}paralleloverline{ZY}$$
            q.e. $ABC$ and $XYZ$ are indeed similar triangles, provided $ABC$ was an isoceles triangle, as asumed by the chosen coordinatisation.



            --- rk






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for the solution, but I like pure geometric proofs (I rarely prefer trig over pure geometry, but at times I’m forced to use trig) more than any other...
              $endgroup$
              – Yellow
              Jan 16 at 17:49
















            1












            $begingroup$

            Let
            $$A=(0, a)\
            B=(-b, 0)\
            C=(b, 0)$$

            and thus
            $$Z=(0, 0)$$
            then we get
            $$tan(angle ABC)=frac ab$$
            and thus
            $$tan(frac 12 angle ABC)=frac{a/b}{1+sqrt{1+a^2/b^2}}=frac a{b+sqrt{a^2+b^2}}$$



            So we can deduce the center $M$ of the incircle to be
            $$M=(0, frac {ab}{b+sqrt{a^2+b^2}})$$



            Now define lines $g$ and $h$ by
            $$g=overline{AC}: y=-frac ab x+a\
            h=overline{ME}: y=frac ba x+frac {ab}{b+sqrt{a^2+b^2}}$$

            Equating those will then provide
            $$frac {a^2+b^2}{ab}cdot x=frac {ab+asqrt{a^2+b^2}-ab}{b+sqrt{a^2+b^2}}$$
            or
            $$x=frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}$$
            Inserting $x$ into $h$ further provides
            $$y=frac{ab^2}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}+frac{ab}{b+sqrt{a^2+b^2}}cdotfrac{sqrt{a^2+b^2}}{sqrt{a^2+b^2}}=frac{ab}{sqrt{a^2+b^2}}$$
            Thus we have
            $$E=(frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})$$



            Now define line $k$ to be
            $$k=overline{BM}: y=frac a{b+sqrt{a^2+b^2}} (x+b)$$
            and intersecting that with $overline{DE}$, i.e. equating it with the $y$ value of $E$, provides
            $$frac a{b+sqrt{a^2+b^2}} (x+b)=frac{ab}{sqrt{a^2+b^2}}\
            x+b=frac{b(b+sqrt{a^2+b^2})}{sqrt{a^2+b^2}}=frac{b^2}{sqrt{a^2+b^2}}+b\
            x=frac{b^2}{sqrt{a^2+b^2}}$$



            Thus we have calculated $Y$ to be
            $$Y=(frac{b^2}{sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})=frac b{sqrt{a^2+b^2}} (b, a)$$
            and this finally proves your conjecture:
            $$overline{AB}paralleloverline{ZY}$$
            q.e. $ABC$ and $XYZ$ are indeed similar triangles, provided $ABC$ was an isoceles triangle, as asumed by the chosen coordinatisation.



            --- rk






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for the solution, but I like pure geometric proofs (I rarely prefer trig over pure geometry, but at times I’m forced to use trig) more than any other...
              $endgroup$
              – Yellow
              Jan 16 at 17:49














            1












            1








            1





            $begingroup$

            Let
            $$A=(0, a)\
            B=(-b, 0)\
            C=(b, 0)$$

            and thus
            $$Z=(0, 0)$$
            then we get
            $$tan(angle ABC)=frac ab$$
            and thus
            $$tan(frac 12 angle ABC)=frac{a/b}{1+sqrt{1+a^2/b^2}}=frac a{b+sqrt{a^2+b^2}}$$



            So we can deduce the center $M$ of the incircle to be
            $$M=(0, frac {ab}{b+sqrt{a^2+b^2}})$$



            Now define lines $g$ and $h$ by
            $$g=overline{AC}: y=-frac ab x+a\
            h=overline{ME}: y=frac ba x+frac {ab}{b+sqrt{a^2+b^2}}$$

            Equating those will then provide
            $$frac {a^2+b^2}{ab}cdot x=frac {ab+asqrt{a^2+b^2}-ab}{b+sqrt{a^2+b^2}}$$
            or
            $$x=frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}$$
            Inserting $x$ into $h$ further provides
            $$y=frac{ab^2}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}+frac{ab}{b+sqrt{a^2+b^2}}cdotfrac{sqrt{a^2+b^2}}{sqrt{a^2+b^2}}=frac{ab}{sqrt{a^2+b^2}}$$
            Thus we have
            $$E=(frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})$$



            Now define line $k$ to be
            $$k=overline{BM}: y=frac a{b+sqrt{a^2+b^2}} (x+b)$$
            and intersecting that with $overline{DE}$, i.e. equating it with the $y$ value of $E$, provides
            $$frac a{b+sqrt{a^2+b^2}} (x+b)=frac{ab}{sqrt{a^2+b^2}}\
            x+b=frac{b(b+sqrt{a^2+b^2})}{sqrt{a^2+b^2}}=frac{b^2}{sqrt{a^2+b^2}}+b\
            x=frac{b^2}{sqrt{a^2+b^2}}$$



            Thus we have calculated $Y$ to be
            $$Y=(frac{b^2}{sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})=frac b{sqrt{a^2+b^2}} (b, a)$$
            and this finally proves your conjecture:
            $$overline{AB}paralleloverline{ZY}$$
            q.e. $ABC$ and $XYZ$ are indeed similar triangles, provided $ABC$ was an isoceles triangle, as asumed by the chosen coordinatisation.



            --- rk






            share|cite|improve this answer











            $endgroup$



            Let
            $$A=(0, a)\
            B=(-b, 0)\
            C=(b, 0)$$

            and thus
            $$Z=(0, 0)$$
            then we get
            $$tan(angle ABC)=frac ab$$
            and thus
            $$tan(frac 12 angle ABC)=frac{a/b}{1+sqrt{1+a^2/b^2}}=frac a{b+sqrt{a^2+b^2}}$$



            So we can deduce the center $M$ of the incircle to be
            $$M=(0, frac {ab}{b+sqrt{a^2+b^2}})$$



            Now define lines $g$ and $h$ by
            $$g=overline{AC}: y=-frac ab x+a\
            h=overline{ME}: y=frac ba x+frac {ab}{b+sqrt{a^2+b^2}}$$

            Equating those will then provide
            $$frac {a^2+b^2}{ab}cdot x=frac {ab+asqrt{a^2+b^2}-ab}{b+sqrt{a^2+b^2}}$$
            or
            $$x=frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}$$
            Inserting $x$ into $h$ further provides
            $$y=frac{ab^2}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}+frac{ab}{b+sqrt{a^2+b^2}}cdotfrac{sqrt{a^2+b^2}}{sqrt{a^2+b^2}}=frac{ab}{sqrt{a^2+b^2}}$$
            Thus we have
            $$E=(frac{a^2b}{(b+sqrt{a^2+b^2}) sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})$$



            Now define line $k$ to be
            $$k=overline{BM}: y=frac a{b+sqrt{a^2+b^2}} (x+b)$$
            and intersecting that with $overline{DE}$, i.e. equating it with the $y$ value of $E$, provides
            $$frac a{b+sqrt{a^2+b^2}} (x+b)=frac{ab}{sqrt{a^2+b^2}}\
            x+b=frac{b(b+sqrt{a^2+b^2})}{sqrt{a^2+b^2}}=frac{b^2}{sqrt{a^2+b^2}}+b\
            x=frac{b^2}{sqrt{a^2+b^2}}$$



            Thus we have calculated $Y$ to be
            $$Y=(frac{b^2}{sqrt{a^2+b^2}}, frac{ab}{sqrt{a^2+b^2}})=frac b{sqrt{a^2+b^2}} (b, a)$$
            and this finally proves your conjecture:
            $$overline{AB}paralleloverline{ZY}$$
            q.e. $ABC$ and $XYZ$ are indeed similar triangles, provided $ABC$ was an isoceles triangle, as asumed by the chosen coordinatisation.



            --- rk







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            edited Jan 12 at 20:01

























            answered Jan 12 at 19:55









            Dr. Richard KlitzingDr. Richard Klitzing

            1,78016




            1,78016












            • $begingroup$
              Thank you for the solution, but I like pure geometric proofs (I rarely prefer trig over pure geometry, but at times I’m forced to use trig) more than any other...
              $endgroup$
              – Yellow
              Jan 16 at 17:49


















            • $begingroup$
              Thank you for the solution, but I like pure geometric proofs (I rarely prefer trig over pure geometry, but at times I’m forced to use trig) more than any other...
              $endgroup$
              – Yellow
              Jan 16 at 17:49
















            $begingroup$
            Thank you for the solution, but I like pure geometric proofs (I rarely prefer trig over pure geometry, but at times I’m forced to use trig) more than any other...
            $endgroup$
            – Yellow
            Jan 16 at 17:49




            $begingroup$
            Thank you for the solution, but I like pure geometric proofs (I rarely prefer trig over pure geometry, but at times I’m forced to use trig) more than any other...
            $endgroup$
            – Yellow
            Jan 16 at 17:49


















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