Let $ABCD$ be a square and $M, N$ points on sides $AB, BC$, respectably, such that $\angle MDN = 45◦$ . If $ R$ is the midpoint of $MN$ show that $RP = RQ$ where $P, Q$ are the points of intersection of $AC$ with the lines $MD, ND.¢$
How do I prove this geometry problem?
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euclidean-geometry
1 Answers
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We have $\angle QAM=\angle QDM=45°$, so quadrilateral $QDAM$ is cyclic, and $\angle AQM=\angle ADM$. Similarly, we have $\angle CPN=\angle CDN$.
Also, using external angle propiety, we have $\angle DPC=\angle DAP+\angle PDA$, and $\angle DQP =\angle QDC+\angle QCD$. This way, $\angle DPN=\angle MQD=45°+\angle ADM+\angle QDC=90°$
Now, triangles $PMN$ and $QMN$ are right, so $\overline{PR}=\overline{MR}=\overline{RN}$ and $\overline{QR}=\overline{MR}=\overline{RN}$ (by a well-known propiety), and $\overline{PR}=\overline{QR}$.