# Show $\triangle AES$ is equilateral iff $\triangle IKE$ is equilateral....

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!

• Hint: uniqueness Commented Oct 3, 2023 at 13:29
• Please use descriptive titles. Commented Oct 3, 2023 at 13:53
• 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 $AS=AE.$ Commented Oct 3, 2023 at 14:00
• Because the OP wants to continue his argument. A new proof does not answer his question. Commented Oct 3, 2023 at 14:04
• @LiKwokKeung Any idea that will lead to the corect answer are good for me. Maybe my idea isnt going anywhere. Commented Oct 3, 2023 at 17:16

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

But we first need a lemma.

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 $$\Delta AES \sim \Delta IKE$$ :

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

• 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 $AS \parallel BC$. Commented Oct 4, 2023 at 2:24

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.

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