Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
Here's the diagram:
I want to use projective geo .
My Progress : It is very known that $P,Q,R$ are collinear[simson line]
Now, here's a lemma.
Lemma: Given a cyclic quad $ABCD$, the angle bisectors $\angle ABC$ and $\angle ADC$ are concurrent with $AC$ if and only if $ABCD$ is harmonic .
Proof : if $ABCD$ is harmonic, then $(A,C;B,D)=-1 \implies \frac {BA}{BC}=\frac {DA}{DC} $ , now applying the angle bisector theorem , we are done.
We can go backwards for proving the other direction.
So, the rephrased question is:
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $Q$ is the midpoint of $PR$ if and only if $ABCD$ is harmonic:
Now , since , I am asked to use Projective geo , I am planning on showing $(P,R;Q,P_{\infty})=-1$ . Now , I can get $P_{\infty}$ when I consider a line parallel to simson line but I am not able to proceed on taking which line. I took a line parallel to $PR$ through $D$, but not able to proceed..
Best Answer
Extend $DQ$ to meet the circle at $X$.Then, BX would be parallel to the simson line as mentioned in the comment.
If part: Now,let $BX$ meet $AC$ at $Y$.Take perspectivity at $B$ onto line $AC$ to show that $(Y,Q,A,C)$ harmonic.Now,take perspectivity at $X$ onto the circle to show that $ABCD$ is harmonic. The other direction can be proved similarly.