$ PC'$ is the angular bisector of $ \angle{APB}$ and so the fact that $ Q$ is the intersection of the bisector of $ AB$ and the external angular bisector of $ \angle{APB}$ would imply $ AQPB$ is incriptible. This shows the first equality. Let $ M$ be the intersection of $ PC'$ and the circle passing through $ A, Q, P, B$. Then $ \angle{PBQ} = \angle{PAQ} = \angle{PMQ}$. But we also have $ \angle{PMQ} = \angle{PCC'}$ (notice that $ MQ // CC'$). This completes the proof.

You forgot to prove this mathangel wrote: $ PC'$ is the angular bisector of $ \angle{APB}$ and so the fact that $ Q$ is the intersection of the bisector of $ AB$ and the external angular bisector of $ \angle{APB}$ would imply $ AQPB$ is incriptible It follows from $ (A,B,C',D)=-1,$ where $ D\equiv{}A'B' \cap AB.$

Let QD be the bisector of the triangle AQB. Let H be the orthocentre of triangle ABC. Let AB intersect A'B' in M. <A'C'B=<ACB, because AA' and CC' are alptitudes. Let BB' intersect DQ at E. Then B'ADE is circumcsribed(because <AB'E=<ADE=90 degrees), so <EB'D=<EAD, but DQ is bisector of AB, so <EAD=<EBD=90degrees-<BAC. So <DB'B=90degrees-<BAC. A'HB'C is circumscribed, so <BB'A'=<C'CB=90degrees-<ABC. <DB'A'=<DB'B+<BB'A'=90degrees-<BAC+90degrees-<ABC=<ACB. Then <A'C'B=<ACB=<DB'A' and A'B'DC' is circumscribed. So MA*MB=MD*MC'=MA'*MB*. <QDM=<C'PM=90 degrees, then triangles MPC' and MDQ are similar. So MD*MC'=MP*MQ. Then MA*MB=MP*MQ and ABPQ is circumscribed. Then <APB'=<ABQ=<BAQ=<BPA'. Then <APC'=<C'PB. PC' is bisector of <APB. QD is bisector of <AQB. Let PC' intersect QD in X. Then AXBQP is circumscrtibed. <CC'B=<QDB, so CC'||QX. So <PC'C=<PXQ=<PBQ=<PAQ, then <PC'C=<PBQ=<PAQ.

Using nine point circle and given 2 circles, if use power of some point, it is easy to show that

khashi70 wrote: $ ABC$ is a triangle and $ AA'$ , $ BB'$ and $ CC'$ are three altitudes of this triangle . Let $ P$ be the feet of perpendicular from $ C'$ to $ A'B'$ , and $ Q$ is a point on $ A'B'$ such that $ QA = QB$ . Prove that : $ \angle PBQ = \angle PAQ = \angle PC'C$ Let I be the intersection of A'B' and AB. We have $ (ABC'I)=-1$ and $ \angle C'PI=90^o$ so $ PC'$ is the bisector of $ \angle APB$ Let $ (ABP) \cap A'B'=\{Q'\}$ $ \Rightarrow \angle BAQ'=\angle BPA'=\angle APQ'=\angle ABQ'$ so $ Q'A=Q'B\Rightarrow Q'\equiv Q$ $ \Rightarrow \angle PBQ=\angle PAQ$ On the other hand, $ \angle PC'C=\angle C'IP=\angle ABP-\angle BPI=\angle ABP-\angle ABQ=\angle QBP$

Quote: the fact that is the intersection of the bisector of and the external angular bisector of would imply is incriptible. why is this the case? :

Let $J$ be the intersection point of $A'B'$ and $AB$ and drop $QX \perp AB$ with $X$ on $AB$.Then note that $X$ is the midpoint of $AB$ and since $AA',BB',CC'$ are concurrent we have $(J,C',A,B)=-1$.So $X$ being the midpoint of $AB$ and noting that $PC'XQ$ is cyclic( $\angle{PC'X}=\angle{PQX}=90^{\circ}$) we have $JA \cdot JB=JC' \cdot JX=JP \cdot JQ$ so $APQB$ is cyclic.So $\angle{PAQ}=\angle{PBQ}$.Next note that if $K=AQ \cap CC'$ then $\angle{AKC}=\angle{AQX}=\frac{1}{2}\angle{AQB}$.Also since $(J,C';A,B)=-1$ we see that $PJ,PC',PA,PB$ form a harmonic pencil.Now since $PC' \perp PJ$,$PC'$ must be the internal angle bisector of $\angle{APB}$.Thus $\angle{APC'}=\frac{1}{2}\angle{APB}=\frac{1}{2}\angle{AQC}$.So $\angle{APC'}=\angle{AKC'}=\frac{1}{2}\angle{AQC} \implies APKC'$ is cyclic.Thus $\angle{PAQ}=\angle{PAK}=\angle{PC'K}=\angle{PC'C}$ as desired.

Nice and elegant. Here's my solution: Let $A'B'$ meet $AB$ at the point $T$.(working in the projective plane we wont bother about $T$ being the point at infinity.) Now, observe that $C',P,Q,M$ are con-cyclic because of the perpendicularity conditions and that $(B,A;C',T)=-1$. This along with the fact that $M$ is the midpoint of $AB$ gives that $TC'.TM=TA.TB=TP.TQ$ which in turn proves that points $B,A,P,Q$ are con-cyclic. Therefore, $\angle PBQ=\angle PAQ$ and since, $Q$ lies on the perpendicular bisector of $AB$, $PQ$ bisects $\angle APB$. Now, $\angle PC'C=\frac{(\angle A' -\angle B')}{2}$ and $\angle PAQ=\frac{(\angle PBA-\angle PAB)}{2}$. Now, the rest is trivial angle-chase.