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Romanian Mathematical Olympiad

Geometry Difficulty 6.2 National Olympiad Prove it Romania

In triangle $ABC$ , we have $m(\angle A) < m(\angle C)$ . Let $E$ be a point on the bisector of angle $B$ such that $\angle EAB = \angle ACB$ . Let $D$ be a point on line $BC$ such that $B \in (CD)$ and $[BD] = [AB]$ . Prove that the midpoint $M$ of the line segment $[AC]$ belongs to the line $DE$ .

Bogdan Antohe

Solution

Let $N$ be the midpoint of $[AD]$ ; it follows that $BN \perp AD$ . If $AP \perp BE$ , $P \in BE$ ; then the quadrilateral $BNAP$ is a rectangle.

Since the points $N$ and $P$ are the projections of $A$ on the interior and exterior bisectors of the angle $ABC$ , respectively, it follows that the line $NP$ contains the midline parallel to $BC$ , and hence $M \in (NP)$ .

Let $\{F\} = AC \cap BE$ ; from $\triangle BAE \sim \triangle BCF$ it follows that $\angle AEB = \angle BFC$ and $\angle AFE = \angle AEB$ , hence the triangle $AEF$ is isosceles, with $AE = AF$ , and we deduce that $P$ is the midpoint of the line segment $[EF]$ .

In the trapezoid $ADFE$ , the points $N$ and $P$ are the midpoints of the bases, therefore the line $NP$ contains the point of intersection of trapezoid's diagonals $[AF]$ and $[DE]$ . Since $\{M\} = AC \cap NP$ , it follows that $M \in DE$ .

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