Problem 4

Let

ABCD

be a cyclic quadrilateral. Let

P,Q,R

be the feet of perpendiculars from

D

to lines

BC,CA,AB

, respectively. Show that

PQ=QR

if and only if the bisectors of angles

ABC

and

ADC

meet on segment

AC

Step 1 of 7: Derive the Simson line

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In plain words

The three perpendicular feet are collinear, and the requested equality is exactly a midpoint condition.

P,Q,R\text{ are collinear, and }PQ=QR\iff Q\text{ is the midpoint of }PR

Detailed analysis

Because $DQ\perp AC$ , $DP\perp BC$ , and $DR\perp AB$ , the quadrilaterals $CQDP$ , $BPDR$ , and $AQDR$ are cyclic. In directed angles modulo $180^\circ$ , these cyclicities and the cyclicity of $ABCD$ give $\angle(PQ,QC)=\angle(PD,DC)=\angle(DR,DA)=\angle(QR,QA)$ . Since $QA$ and $QC$ are the same line, $PQ$ and $QR$ are the same line. Thus $P,Q,R$ are collinear, and on this line $PQ=QR$ exactly when $Q$ is the midpoint of $PR$ .