angle chasing , <ECB = 30^o, <FBC = 15^o, right isosceles

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angle chasing , <ECB = 30^o, <FBC = 15^o, right isosceles

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Source: 2020 Argentina OMA L3 p3

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#1 h Dec 26, 2020, 4:00 AM • 1 Y

Y by Mango247

Let $ABC$ be a right isosceles triangle with right angle at $A$ . Let $E$ and $F$ be points on A $B$ and $AC$ respectively such that $\angle ECB = 30^o$ and $\angle FBC = 15^o$ . Lines $CE$ and $BF$ intersect at $P$ and line $AP$ intersects side $BC$ at $D$ . Calculate the measure of angle $\angle FDC$ .

This post has been edited 1 time. Last edited by parmenides51, Dec 26, 2020, 4:01 AM

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L567

#2 h Dec 26, 2020, 7:09 AM

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Let $AB = AC = a$ and $BC = \sqrt{2}a$

By trig, we get that $AF = \frac{a}{\sqrt{3}}$ , $FC = \frac{(\sqrt{3} - 1)a}{\sqrt{3}}$ , $AE = (2 - \sqrt{3})a$ and $BE = (\sqrt{3} - 1)a$

Now, by Ceva's theorem, we get that $\frac{CD}{DB} = 2 - \sqrt{3}$ . Since $CD + DC = \sqrt{2}a$ , solving, we get that $CD = \frac{(3\sqrt{2} - \sqrt{6})a}{6}$ and $DB = \frac{(3\sqrt{2} + \sqrt{6})a}{6}$

Now, observe that $CD.CB = CF.CA$ , which implies that $BDFA$ is cyclic. So, $\angle FDC = \angle FAB = 90^\circ$

This post has been edited 1 time. Last edited by L567, Dec 26, 2020, 7:10 AM

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