AF_|_ CE wanted, <ACD=<DCE =< ECB, AB=AC, 72^o (2019 Moldova NMO 7.2)

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AF_|_ CE wanted, <ACD=<DCE =< ECB, AB=AC, 72^o (2019 Moldova NMO 7.2)

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#1 h Mar 18, 2021, 2:12 AM

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Let $ABC$ be an isosceles triangle with $AB=AC$ and $\angle BAC= 72^o$ . On the side $AB$ the points $D$ and $E$ are taken so that $\angle ACD = \angle DCE =\angle ECB$ , and the point $F$ belongs to the side $(BC)$ , so that $EF$ is the bisector of the angle $BEC$ . Prove that $AF \perp CE$ .

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#2 h Jan 16, 2025, 5:26 PM

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Given $\triangle ABC\ :\ A(0,\tan 54^{\circ}),B(-1,0),C(1,0)$ .

The line $EC\ :\ y=-\tan 18^{\circ}(x-1)$ intersects the line $AB\ :\ y=\tan 54^{\circ}(x+1)$ in ther point $E(-\frac{\sin 36^{\circ}}{\sin 72^{\circ}},\frac{2\sin 18^{\circ}\sin 54^{\circ}}{\sin 72^{\circ}})$ .
The line $EF\ :\ y-\frac{2\sin 18^{\circ}\sin 54^{\circ}}{\sin 72^{\circ}}=-\tan 72^{\circ}(x+\frac{\sin 36^{\circ}}{\sin 72^{\circ}})$ intersects $BC$ in the point $F(-\frac{\sin 18^{\circ}}{2\sin^{2}36^{\circ}},0)$ .

The slope of the line $AF\ :\ m_{AF}=\frac{1}{tan 18^{\circ}}$ , so $AF \bot EC$ .

This post has been edited 1 time. Last edited by vanstraelen, Jan 16, 2025, 8:22 PM

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#3 h Jan 16, 2025, 8:03 PM

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By simple angle chase, $\angle{CFE} = 108$ . Thus, $\angle{BAC} + \angle{CFE} = 180$ and $ACFE$ is cyclic. Because $ACFE$ is cyclic, $\angle{CEF} = \angle{CAF} = 54$ . Then $\angle{ACE} + \angle{CAF} = 90$ , thus $AF \perp CE$ .

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