Track / Stage 6 / 157 of 400 #1157 of 1964

Problem 1157

National olympiad, first round

Geometry Difficulty 6.2 Prove it

4. The perpendicular bisectors of sides $AB$ and $BC$ of a convex quadrilateral $ABCD$ intersect sides $CD$ and $DA$ at points $P$ and $Q$ respectively. It turns out that $\cdot A P B=\cdot B Q C$ . Inside the quadrilateral, a point $X$ is chosen such that $Q X \| A B$ and $P X \| B C$ . Prove that the line $B X$ bisects the diagonal $AC$ . (S. Berlov)

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

Solution. It is sufficient to prove that the distances from points $A$ and $C$ to the line $B X$ are equal. This is equivalent to $S_{A B X}=S_{B C X}$ , since triangles $A B X$ and $B C X$ share the same base $B X$ . Since $Q X \| A B$ , we have $\quad S_{A B X}=S_{A B Q}$ . Similarly, $\quad S_{C B X}=S_{C B P}$ . Note that isosceles triangles $A B P$ and $C B Q$ are similar by two angles,

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therefore $A B / B C=B P / B Q$ , from which $A B \cdot B Q=C B \cdot B P$ . Since $\cdot A B P=\cdot C B Q$ , then $\cdot A B Q=\cdot C B P$ . Consequently, the areas of triangles $A B Q$ and $C B P$ are proportional to the products of the sides enclosing equal angles, i.e., these areas are equal. Thus, $S_{A B X}=S_{A B Q}=S_{C B P}=S_{C B X}$ , which is what we needed to prove.

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