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Brazilian Math Olympiad

Geometry Difficulty 4.5 AIME Prove it Brazil

Let $ABCD$ be a convex quadrilateral such that $AD = DC$ , $AC = AB$ and $\angle ADC = \angle CAB$ . Let $M$ and $N$ be the midpoints of $AD$ and $AB$ . Prove that triangle $MNC$ is isosceles.

Solution

Since $AD = CD$ , $AB = AC$ and $\angle ADC = \angle BAC$ , triangles $ADC$ and $BAC$ are similar by case SAS. Segments $CM$ and $CN$ are corresponding medians, so $\frac{CM}{CN} = \frac{CA}{CB}$ and $\angle$ BCN = $\angle$ ACM $\iff \angle$ BCN + $\angle$ NCA = $\angle$ ACM + $\angle$ NCA $\iff \angle$ BCA = $\angle$ NCM. Thus, again by case SAS, triangles $CMN$ and $CAB$ are similar, and therefore $CMN$ is an isosceles triangle with $CM = MN$ .

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