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 3 of 7: Construct the parallel chord

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In plain words

A second intersection on DQ produces a chord parallel to the Simson line.

\angle DQR=\angle DAR=\angle DAB=\angle DEB,\qquad BE\parallel PQR

Detailed analysis

Let $E\ne D$ be the second intersection of $DQ$ with the circumcircle, and let $F\ne B$ be the second intersection of $BQ$ with it. The cyclic quadrilateral $AQDR$ gives $\angle DQR=\angle DAR$ ; since $A,R,B$ are collinear, $\angle DAR=\angle DAB$ . Since $A,B,D,E$ are concyclic, $\angle DAB=\angle DEB$ . As $D,E,Q$ are collinear, the last equality says that $BE$ is parallel to $QR$ , hence to the Simson line $PQR$ .