AoPS - Problem

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Problem

Source: 2020 Argentina OMA L3 p3

Tags: geometry, angles, right triangle, isosceles

parmenides51

26.12.2020 07:00

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 $\angle FDC$ .

L567

26.12.2020 10:09

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, $\angle FDC = \angle FAB = 90^\circ$