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Problem 1105

AMC 12 late, AIME early

Geometry Difficulty 4.8 Prove it Berkeley Math Circle · United States

Two congruent line segments $AB$ and $CD$ intersect at a point $E$ . The perpendicular bisectors of $AC$ and $BD$ intersect at a point $F$ in the interior of $\angle AEC$ . Prove that $EF$ bisects $\angle AEC$ .

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:

Since $F$ is on the perpendicular bisectors of $AC$ and $BD$ , $FA = FC$ and $FB = FD$ . Also $AB = CD$ is given. Thus $\triangle FAB \cong \triangle FCD$ by SSS. Since corresponding heights of congruent triangles are equal, $F$ is equidistant from $AB$ and $CD$ , implying that $F$ is on the bisector of $\angle AEC$ .

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