Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.2 Shortlist Prove it China

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 $AC$ .

Solution

Proof By Simson's Theorem, we know that $P$ , $Q$ , $R$ are collinear. Moreover, since $\angle DPC$ and $\angle DQC$ are right angles, the points $D$ , $P$ , $Q$ , $C$ are concyclic and so $\angle DCA = \angle DPQ = \angle DPR$ . Similarly, since $D$ , $Q$ , $R$ , $A$ are concyclic, we have $\angle DAC = \angle DRP$ . Therefore $\triangle DCA \sim \triangle DPR$ .

Likewise, $\triangle DAB \sim \triangle DQP$ and $\triangle DBC \sim \triangle DRQ$ . Then
$\frac{DA}{DC} = \frac{DR}{DP} = \frac{DB \cdot \frac{QR}{BC}}{DB \cdot \frac{PQ}{BA}} = \frac{QR}{PQ} \cdot \frac{BA}{BC}.$
Thus $PQ = QR$ if and only if $\frac{DA}{DC} = \frac{BA}{BC}$ .

Now the bisectors of the angles $ABC$ and $ADC$ divide $AC$ in the ratios of $\frac{BA}{BC}$ and $\frac{DA}{DC}$ respectively. This completes the proof.

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