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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

enter image description here

I don't know what to do forward. I got stuck with that $60°$ angle. Hope one of you can help me! Thank you!

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  • $\begingroup$ Hint: uniqueness $\endgroup$ Commented Oct 3, 2023 at 13:29
  • $\begingroup$ Please use descriptive titles. $\endgroup$ Commented Oct 3, 2023 at 13:53
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    $\begingroup$ If we can prove that $\angle FSE=\frac{\angle A}{2}$. Then the circle with $A$ as center, $AE=AF$ as radius will pass through $S$. This implies $AS=AE.$ $\endgroup$ Commented Oct 3, 2023 at 14:00
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    $\begingroup$ Because the OP wants to continue his argument. A new proof does not answer his question. $\endgroup$ Commented Oct 3, 2023 at 14:04
  • $\begingroup$ @LiKwokKeung Any idea that will lead to the corect answer are good for me. Maybe my idea isnt going anywhere. $\endgroup$ Commented Oct 3, 2023 at 17:16

2 Answers 2

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I would prove that $\Delta AES \sim \Delta IKE$ and consequently one is equilateral iff the other is.

But we first need a lemma.

enter image description here In the figure, if $AX=AY$ and $\angle XAY = 2 \times \angle XZY$, then $Z$ lies on the circle with $A$ as center, $AX$ as radius. In other words, $AX=AY=AZ.$

Reason: $\mathrm{angle \; at \; center} = 2 \times \mathrm{angle \; at \; circumference.}$

Proof that $\Delta AES \sim \Delta IKE$ : enter image description here

$(1)$ $I, F, B, D$ 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.

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  • $\begingroup$ Surprise use of a basic lemma, nice. So $\triangle CDE \sim \triangle IEK \sim \triangle AES$ because their angles are $C, 90- C/2, 90- C/2$. Also $AS \parallel BC$. $\endgroup$ Commented Oct 4, 2023 at 2:24
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enter image description here

Hint:You showed $\angle AES=60^o$. Extend DK to meet AS or its extension at H. A perpendicular from A on BC meets it at P. We have:

$\angle EDC=\angle DEC=60^o$

$\Rightarrow \angle ACP= 60^o$

$\Rightarrow \angle PAC=30^o$

Also $\angle SDH=30^o\Rightarrow HS||BC\Rightarrow \angle ASE=60^o$

that is triangle AES is equilateral.

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  • $\begingroup$ Thank you so much for your answer! May you explain me why $HS$ and $BC$ are parallel??? $\endgroup$ Commented Oct 3, 2023 at 18:33
  • $\begingroup$ @lonela buciu, You'r welcome, because they are both perpendicular on BC. $\endgroup$
    – sirous
    Commented Oct 4, 2023 at 5:11

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