Prove that triangle $XYZ$ is equilateral 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.
 A: 
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:
$$\angle BXC=\angle BXI=\angle BDI=90^\circ\tag{1}$$
In a similar way we can show that:
$$\angle BYC=90^\circ\tag{2}$$
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:
$$\angle XZY=2\angle XBY=2(\angle XBC-\angle IBC)=2(90^\circ-\gamma/2-\beta/2)=\alpha$$
So triangle $XYZ$ is equilateral if and only if $\alpha=60^\circ$.
A: 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
$$\frac {a^2+b^2}{ab}\cdot x=\frac {ab+a\sqrt{a^2+b^2}-ab}{b+\sqrt{a^2+b^2}}$$
or
$$x=\frac{a^2b}{(b+\sqrt{a^2+b^2})\ \sqrt{a^2+b^2}}$$
Inserting $x$ into $h$ further provides
$$y=\frac{ab^2}{(b+\sqrt{a^2+b^2})\ \sqrt{a^2+b^2}}+\frac{ab}{b+\sqrt{a^2+b^2}}\cdot\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{a^2+b^2}}$$
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:
$$\overline{AB}\parallel\overline{ZY}$$
q.e. $ABC$ and $XYZ$ are indeed similar triangles, provided $ABC$ was an isoceles triangle, as asumed by the chosen coordinatisation.
--- rk
