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

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 $$\alpha=60^\circ$$.

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

• 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... – Yellow Jan 16 at 17:49