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Romanian Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Romania

The convex quadrilateral $ABCD$ has $\angle BCD = \angle ADC \ge 90^\circ$ . The bisectors of the angles $\angle BAD$ and $\angle ABC$ meet at a point $M$ , placed on the line $CD$ . Prove that $M$ is the midpoint of the segment $[CD]$ .

Solution

Case I: $AD$ and $BC$ have a common point $E$ . Then $M$ is the incenter of the triangle $ABE$ , hence $(EM)$ is the bisector of the angle $\angle AEB$ .
Since $\angle ECD = \angle EDC$ , the triangle $EDC$ is isosceles with base $[DC]$ . Therefore $*$ is a median in triangle $EDC$ , so $M$ is the midpoint of the segment $CD$ .

Case II: $AD \parallel BC$ . Then $\angle CAB = \angle ABD = 90^\circ$ , hence $\angle MAB + \angle MBA = 90^\circ$ , that is triangle $MAB$ has a right angle in $M$ .
Denote $N$ the midpoint of the segment $[AB]$ . Then triangle $NAM$ is isosceles with base $[AM]$ , so $\angle NMA = \angle NAM = \angle MAD$ , hence $MN \parallel AD$ . It follows that $MN$ is the central median of the trapezoid $ABCD$ , whence the conclusion.

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