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Problem 985

AMC 12 late, AIME early

Geometry Difficulty 4.5 Prove it Irish Mathematical Olympiad · Ireland

Assume $ABCD$ is a convex quadrilateral such that the triangles $ABD$ , $BCD$ , $CDA$ , and $ABC$ have the same area. Prove that $ABCD$ is a parallelogram.

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

Let $B'$ and $D'$ on $AC$ be the feet of the altitudes of the triangles $\triangle ABC$ and $\triangle ADC$ . Because these two triangles have the same area, we get $|BB'| = |DD'|$ . This implies that the two right triangles $BMB'$ and $DMD'$ , which have equal angles at $M$ , are congruent. In particular, $|BM| = |MD|$ .

A similar argument, using the other diagonal, shows that $|AM| = |MC|$ . Because the two triangles $DMA$ and $BMC$ have equal angles at $M$ , they are now seen to be congruent. This shows that $\angle MBC = \angle MDA$ , hence $BC$ is parallel to $DA$ . In a similar way it follows that $AB$ is parallel to $CD$ , hence $ABCD$ is a parallelogram.

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