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Romanian Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Romania

Consider an isosceles trapezoid $ABCD$ with perpendicular diagonals. The parallel from the intersection point of the diagonals meets the non-parallel sides $[BC]$ and $[AD]$ at points $P$ and $R$ respectively. Point $Q$ is the mirror image of $P$ across the midpoint of $[BC]$ . Show that

a) $QR = AD$ ;

b) $QR \perp AD$ .

Solution

Let the diagonals $AC$ and $BD$ meet at point $O$ and let $M$ be the midpoint of the line segment $[BC]$ .

a) Since $OM$ joins the midpoints of two sides of the triangle $PQR$ , $MO \parallel RQ$ and $OM = \frac{RQ}{2}$ . On the other hand, $[OM]$ is a median of the right-angled triangle $BOC$ , hence $OM = \frac{1}{2}BC = \frac{1}{2}AD$ . Consequently, $RQ = AD$ .

b) Let the lines $MO$ and $AD$ meet at $T$ . Then $\angle MBO = \angle MOB = \angle DOT$ . Since $\angle OCB = \angle TDO$ , it follows that $\angle OTD = \angle BOC = 90^\circ$ , so $MT \perp AD$ .

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