Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it

Example 7 As shown in Figure 7, $P$ is any point on the midline $M N$ of $\triangle A B C$ , and the extensions of $B P$ and $C P$ intersect $A C$ and $A B$ at points $D$ and $E$ respectively. Prove: $\frac{A D}{D C}+\frac{A E}{E B}=1$ .

Solution

Prove as shown in Figure 7, draw a line through point $A$ parallel to $BC$ , intersecting the extensions of $CP$ and $BP$ at points $F$ and $G$ respectively.
Since $FG \parallel BC$ , we have
$\begin{array}{l} \frac{AD}{DC}=\frac{AG}{BC}, \frac{AE}{EB}=\frac{AF}{BC}. \\ \text{Therefore, } \frac{AD}{DC}+\frac{AE}{EB} \\ =\frac{AG}{BC}+\frac{AF}{BC}=\frac{FG}{BC}. \end{array}$

It is easy to see that $BP = PG, CP = PF$ .
Thus, quadrilateral $BCGF$ is a parallelogram.
Therefore, $FG = BC$ .
Hence, $\frac{AD}{DC}+\frac{AE}{EB}=1$ .
[Summary] Adding parallel lines and substituting ratios is a common method for solving problems related to "ratios".

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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.