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Problem Of The Week #382 Sep 5th, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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Let $ABCD$ be an inscribed quadrilateral. Let $P$, $Q$ and $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$ and $AB$ respectively. Show that $PQ = QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ meet on $AC.$

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
No one answered last week's POTW. (Sadface) You can refer to the suggested solution as shown below.

potw 382.png

Denote in the triangle $ABC$ the angles at $A$ and $C$ by $\alpha$ and $\gamma$ respectively. $P$ and $Q$ are incident to the Thales circle of $CD$ and $\sin \angle PCQ=\sin \gamma$ or $PQ=CD\sin \gamma$.

Similarly, $QR=AD\sin \alpha$. By condition

$PQ=CD \sin \gamma=AD\sin \alpha=QR$

Rearranging and applying the sine rule for the triangle $ABC$ yields

$\dfrac{CD}{AD}=\dfrac{\sin \alpha}{\sin \gamma}=\dfrac{CB}{AB}$

By the angle bisector theorem, this holds if and only if the bisectors of $\angle ABC$ and $\angle ADC$ meet the segment $AC$ at the very same point.
 
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