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Macedonian Mathematical Competitions

Geometry Difficulty 6.1 National Olympiad Prove it North Macedonia

Two lines $p$ and $q$ intersect at $C$ . The point $C$ divides the line $p$ into two half-lines that, together with one of the half-lines of $q$ with starting point at $C$ , form two angles $pCq$ and $qCp$ . On the bisector of the angle $pCq$ a point $M$ is chosen such that $MN \parallel p$ . The segment $MN$ intersects the line $q$ at a point $D$ . Prove that $D$ is the midpoint of the segment $MN$ .

Solution

Let $P$ be a point on the half-line $Cp$ from the angle $pCq$ and $Q$ be a point on the half-line $Cp$ of the angle $qCp$ . Let $CM$ be the bisector of the angle $pCq$ and $CN$ be the bisector of the angle $qCp$ . $MN \parallel PQ$ and let $D$ be the intersection point of the line $q$ with the line $MN$ .

Because $CM$ is a bisector of the angle $pCq$ and $CN$ is a bisector of the angle $qCp$ , we have that $\angle PCM = \angle MCD$ and $\angle DCN = \angle NCQ$ . $\angle PCM = \angle CMD$ and $\angle NCQ = \angle CND$ , hence $\angle MCD = \angle CMD$ and $\angle DCN = \angle DNC$ .

Hence we get that the triangles $\Delta CMD$ and $\Delta CND$ are isosceles with bases $MC$ and $NC$ correspondingly. So we obtain that $\overline{MD} = \overline{DN}$ , hence $D$ is the midpoint of the segment $MN$ .

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