Track / Stage 6 / 53 of 400 #1053 of 1964

Problem 1053

National olympiad, first round

Geometry Difficulty 6.0 Prove it

$\text{* Problem}$ 3 - 131223}

In any convex quadrilateral $A B C D$ , let $E$ be the midpoint of side $A B$ and $F$ be the midpoint of side $C D$ . The intersection of $A F$ with $D E$ is called $G$ , and the intersection of $B F$ with $C E$ is called $H$ .

It is to be proven that the area of quadrilateral $E H F G$ is equal to the sum of the areas of triangles $A G D$ and $B H C$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

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With "amounts" we denote in the following the area of a polygon. Then we can equivalently rephrase the claim (by adding the four white sub-triangles):

$\begin{aligned} & |E H F G|=|A G D|+|B H C| \\ \Longleftrightarrow & 2|E H F G|+|A E G|+|D F G|+|B E H|+|C F H| \\ \quad= & 2|A G D|+2|B H C|+|A E G|+|D F G|+|B E H|+|C F H| \\ \Longleftrightarrow & |A B F|+|C D E|=(|A E D|+|E B C|)+(|C F B|+|F D A|) \end{aligned}$

Since $F$ is the midpoint of the segment $C D$ , the height of the triangle $A B F$ is exactly the arithmetic mean of the heights of the triangles $A E D$ and $E B C$ . This implies $|A B F|=|A E D|+|E B C|$ and similarly $|C D E|=|C F B|+|F D A|$ , thus proving the claim.

$\text{* Source}$ anonymous}

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