Singapur

Geometry Difficulty 4.5 AIME Prove it Singapore

Let $D$ be the point in the interior of a triangle $ABC$ such that $AB = ab$ , $AC = ac$ , $AD = ad$ , $BC = bc$ , $BD = bd$ and $CD = cd$ . Prove that $\angle ABD + \angle ACD = 60^\circ$ .

Solution

Let $A_1$ , $B_1$ , $C_1$ be the feet of the perpendiculars from $D$ onto the sides $BC$ , $CA$ , $AB$ , respectively.

Then, $B_1C_1 = DA \sin A$ , $C_1A_1 = DB \sin B$ , $A_1B_1 = DC \sin C$ . Thus, $B_1C_1 : C_1A_1 : A_1B_1 = (ad)(bc) : (bd)(ac) : (cd)(ab) = 1 : 1 : 1$ . Therefore, $A_1B_1C_1$ is an equilateral triangle. Consequently, $\angle ABD + \angle ACD = \angle C_1A_1D + \angle B_1A_1D = \angle B_1A_1C_1 = 60^\circ$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.