Dtyjlui

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.

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 $alpha=60^circ$.

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