Question
Let the $\triangle ABC$ and $D$, $E$, $F$ be the intersection points of the center circle $I$, inscribed in the triangle, with the sides $BC$,$ AC$, $AB$, respectively. We denote by $K$ the symmetry of $D$ with respect to the center of the inscribed circle. We denote by $S$ the intersection between $FK$ and $DE$. Show that the $\triangle AES$ is equilateral if and only if the $\triangle IKE$ is equilateral.
My Idea
I don't know what to do forward. I got stuck with that $60°$ angle. Hope one of you can help me! Thank you!
Best Answer
I would prove that $\Delta AES \sim \Delta IKE$ and consequently one is equilateral iff the other is.
But we first need a lemma.
Reason: $\mathrm{angle \; at \; center} = 2 \times \mathrm{angle \; at \; circumference.}$
Proof that $\Delta AES \sim \Delta IKE$ :
$(1)$ $I, F, B, D$ are concyclic $\implies \angle KIF=\angle B \implies \angle IDF=\frac{\angle B}{2}$
$(2)$ $I, D, C, E$ are concyclic $\implies \angle EIK=\angle C \implies \angle IDE=\frac{\angle C}{2}$
$(3)$ In $\Delta DFS, \angle FSD=180^{\text o}-90^{\text o}-\frac{\angle B}{2}-\frac{\angle C}{2}=\frac{\angle A}{2}$
$(4)$ Note that $AE=AF$ by tangent property. Hence by our lemma, $AS=AE$.
$(5)$ Thus both $\Delta AES$ and $\Delta IKE$ are isosceles triangles. We are done if we can prove that $\angle AES = \angle IEK$.
$(6)$ By angle in alternate segment, $\angle AEK=\angle EDK=\frac{\angle C}{2}$.
$(7)$ $\therefore \angle AES=90^{\text o}-\frac{\angle C}{2}$.
$(8)$ $\because IE=IK, \angle IEK=\frac{180^{\text o}-\angle KIE}{2}= \frac{180^{\text o}-\angle C}{2}=90^{\text o}-\frac{\angle C}{2}$.
$(9)$ From $(7)$ and $(8)$, we have $\angle AES=\angle IEK$ .
$(10)$ Thus $\Delta AES \sim \Delta IEK$ and one triangle is equilateral iff the other is equilateral.