Olympiad Maths Prep

Track / Stage 5 / 276 of 400 #876 of 2000

Problem 876

AIME late
Geometry Difficulty 5.7 Prove it

\section*{Problem 2 - 241022}

For a triangle ABCA B C, it is assumed that it is not obtuse and that for the height CDC D perpendicular to ABA B, the equation CDAC=ADBCC D \cdot A C = A D \cdot B C holds.

Prove that these assumptions uniquely determine the size γ\gamma of the interior angle ACB\angle A C B! Determine this angle size γ\gamma!

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

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According to the condition, the triangles ADCA D C and DBCD B C are similar, as they match in the ratio of two sides and the angle opposite the larger side. Therefore, γ=α+β=90\gamma=\alpha+\beta=90^{\circ}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.