Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it

II. (Full marks 25 points) As shown in the figure, in $\triangle ABC$ , $AB=AC$ , $AD$ is the altitude, $E$ is a point on $AB$ , $CF \perp BC$ intersects the extension of $ED$ at $F$ , $M$ and $N$ are the midpoints of $DE$ and $DF$ respectively. Prove: $\angle MAD = \angle NAD$ .

Solution

$\begin{array}{l} \quad \because AB=AC, AD \perp \\ BC, CF \perp BC, \\ \therefore BD=DC, AD \parallel \\ CF. \\ \text { Hence } SD=DT . \end{array}$
$\because M, N$ are the midpoints of $DE, DF$ respectively,
$\begin{array}{l} \therefore SM \parallel AB, NT \parallel CF . \\ \therefore AP=PD, NQ \parallel PD . \\ \therefore PS=PQ, \frac{MD}{DN}=\frac{MP}{PQ} . \end{array}$

From $\frac{PA}{PD}=\frac{PQ}{PS}=1$ , we know that $AQ \parallel BC$ , i.e., $AQ \perp AD$ .
Draw $PR \perp AD$ , intersecting $AM$ at point $R$ , and connect $RD$ . Then $PR \parallel AQ, RA=RD$ .
$\begin{array}{l} \therefore \angle MAD=\angle ADR, \frac{MR}{RA}=\frac{MP}{PQ} . \\ \text { Hence } \frac{MD}{DN}=\frac{MR}{RA} . \\ \therefore RD \parallel AN . \end{array}$

Thus, $\angle ADR=\angle NAD$ .
Therefore, $\angle MAD=\angle NAD$ .

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