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, 2015

Geometry Difficulty 4.3 AIME Prove it Singapore

In an acute-angled triangle $ABC$ , $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$ , $E$ and $F$ are the feet of the perpendiculars from $D$ onto $AB$ and $AC$ respectively. The segments $BF$ and $CE$ intersect at $K$ . Prove that $AK$ is perpendicular to $BC$ .

Solution

Let the extension of $AK$ intersect $BC$ at $N$ . Note that $AE = AF$ and $BD : DC = c : b$ , where $b = AC$ and $c = AB$ . The cevians $AN$ , $BF$ , $CE$ concur at $K$ . By Ceva's theorem, we have $(BN/NC)(CF/FA)(AE/EB) = 1$ . Thus $BN/NC = EB/CF$ . On the other hand, $EB = BD \cos B$ and $CF = DC \cos C$ so that $EB/CF = (BD \cos B)/(DC \cos C) = (c \cos B)/(b \cos C)$ . Therefore, $BN/NC = (c \cos B)/(b \cos C)$ . This shows that $N$ is the foot of the perpendicular from $A$ onto $BC$ .

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