Hopefully correct tenplusten wrote: Given triangle $\triangle ABC $ with $AC\neq BC $,and let $D $ be a point inside triangle such that $\measuredangle ADB=90^{\circ} + \frac {1}{2}\measuredangle ACB $.Tangents from $C $ to the circumcircles of $\triangle ABC $ and $\triangle ADC $ intersect $AB $ and $AD $ at $P $ and $Q $ , respectively.Prove that $PQ $ bisects the angle $\measuredangle BPC $. Solution. Let $E$ be the $C-$ antipode in $\odot(ADC)$. Animate $E$ linearly on line perpendicular to $AC$ through $A$. Therefore $D$ varies with degree $2$, giving degree of $AD$ as $1$ (there is a case when $B=D$). Further point of infinity on the line $CD$ varies with degree $1$ (as degree of $E$ is $1$). Therefore $Q$ has degree $1$. Now for $D = C$, $Q = P$, so locus of $Q$ is line through $P$ which means we need to check for one case of $Q$. For $D = A$, $QA$ is tangent to $\odot(BIC)$ and $QA,QC$ are tangents to circle so $QA = QC$. Therefore $\angle QAB = 90^{\circ} - C/2$, so $2\angle CAD = |2A - 180^{\circ} + C| = |A - B| = \angle CPB$, so we are done.

Radical center of $(ADB),(ADC)$ and point $C$ killed this problem

(ADB) center is midpoint of arc AB (M) And PQ is Radical axis of C and (ADB) BPC Angle bisect line is perpendicular to CM finish

Let $CN$ be angle bisector of $\angle ACB$ with $LN$ on $AB$. Note that $PN=PC$ . Hence angle bisector of angle $\angle BPC$ is $\perp$ bisector of $CN$. So all we need to prove is that $Q$ lies on $\perp$ bisector of $CN$ i.e. we need to prove that $QN=QC$. Now note that $D$ lies on $\odot (AIB)$ where $I$ is incentre of $\triangle ABC$. But $\odot (AIB)$ is $A-$ Apollonius circle of $\triangle CNA\implies D$ lies on $A-$ Apollonius circle of $\triangle CNA$. Hence $N$ lies on $C-$ Apollonius Circle of $\triangle CDA$. But $Q$ is centre of $C-$ Apollonius Circle of $\triangle CDA$. Thus $QN=QC$ as desired.

tenplusten wrote: Given triangle $\triangle ABC $ with $AC\neq BC $,and let $D $ be a point inside triangle such that $\measuredangle ADB=90^{\circ} + \frac {1}{2}\measuredangle ACB $.Tangents from $C $ to the circumcircles of $\triangle ABC $ and $\triangle ADC $ intersect $AB $ and $AD $ at $P $ and $Q $ , respectively.Prove that $PQ $ bisects the angle $\measuredangle BPC $. INMO Test 2: Nice problem but perhaps too easy for Serbian TST. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -35.81762545676626, xmax = 52.691163243213566, ymin = -24.10598715199585, ymax = 29.86278644555289; /* image dimensions */ pen qqwqxt = rgb(0,0.3764705882352941,0.45098039215686275); pen zfqqqq = rgb(0.6235294117647059,0,0); pen qqwwtt = rgb(0,0.4,0.2); pen ccqqqq = rgb(0.8,0,0); pen ffqqtt = rgb(1,0,0.2); pen wwzzqq = rgb(0.4,0.6,0); pen ttzzqq = rgb(0.2,0.6,0); draw((-6.18335851158536,14.094048764439323)--(-9.406991037039226,-0.37189665758930057)--(4.869465956988161,-4.524261515877775)--cycle, linewidth(0.4) + qqwqxt); /* draw figures */ draw((-6.18335851158536,14.094048764439323)--(-9.406991037039226,-0.37189665758930057), linewidth(0.4) + qqwqxt); draw((-9.406991037039226,-0.37189665758930057)--(4.869465956988161,-4.524261515877775), linewidth(0.4) + qqwqxt); draw((4.869465956988161,-4.524261515877775)--(-6.18335851158536,14.094048764439323), linewidth(0.4) + qqwqxt); draw(circle((-0.062357880965784915,5.13787308787178), 10.848028920941598), linewidth(0.4) + zfqqqq); draw(circle((3.7330491577557527,7.3910296393621815), 11.96936115487026), linewidth(0.8) + qqwwtt); draw((-9.406991037039226,-0.37189665758930057)--(-12.283368948955403,-13.279548369860594), linewidth(0.4) + ccqqqq); draw((-12.283368948955403,-13.279548369860594)--(4.869465956988161,-4.524261515877775), linewidth(0.4) + ffqqtt); draw((-6.328946616563997,-5.592306266737558)--(-6.18335851158536,14.094048764439323), linewidth(0.4) + wwzzqq); draw((-6.328946616563997,-5.592306266737558)--(4.869465956988161,-4.524261515877775), linewidth(0.4) + ttzzqq); /* dots and labels */ dot((-6.18335851158536,14.094048764439323),linewidth(4pt) + dotstyle); label("$A$", (-5.903505234124996,14.674431590242744), NE * labelscalefactor); dot((-9.406991037039226,-0.37189665758930057),linewidth(4pt) + dotstyle); label("$B$", (-9.064533401981418,0.25705921489758027), NE * labelscalefactor); dot((4.869465956988161,-4.524261515877775),linewidth(4pt) + dotstyle); label("$C$", (5.19864247737073,-3.9062461769133225), NE * labelscalefactor); dot((-4.352191517478712,2.6186959285510913),linewidth(4pt) + dotstyle); label("$I$", (-4.053147282209042,3.26389088676101), NE * labelscalefactor); dot((-10.650669896783175,7.497402728559194),linewidth(4pt) + dotstyle); label("$J$", (-10.375203617921887,8.121080510540397), NE * labelscalefactor); dot((-6.281411225892663,0.835406886331497),linewidth(4pt) + dotstyle); label("$D$", (-5.980603482121494,1.4906311828415515), NE * labelscalefactor); dot((-6.328946616563997,-5.592306266737558),linewidth(4pt) + dotstyle); label("$Q$", (-6.0577017301179925,-4.9856216488642975), NE * labelscalefactor); dot((-12.283368948955403,-13.279548369860594),linewidth(4pt) + dotstyle); label("$P$", (-11.994266825848346,-12.695446448514117), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We see that $\angle ADB = \angle AIB = 90^\circ + \frac{1}{2}\angle ACB$, which means that the circumcenter of $\triangle ADB$ is $J$, the midpoint of arc $AB$ not containing $C$. The point $P$ is fixed, we would like to show that point $Q$ lies on a fixed line. We see that $P$ is the center of $C$-Apollonius circle of $\triangle ABC$, therefore angle bisector of $\angle BPC$ must be perpendicular to line $\overline{CJ}$. Let $\Gamma$, $\Gamma_B$ and $\Gamma_C$ be the degenerate circle point $C$ and the circumcircles of $\triangle ADB$ and $\triangle ADC$ respectively. Then, the radical axis of $\Gamma$ and $\Gamma C$ is tangent from $C$ to $\Gamma_C$, the radical axis of $\Gamma _B$ and $\Gamma _C$ is line $\overline{AD}$, which intersects with the tangent from $C$ to $\Gamma_C$ at the point $Q$, therefore, $Q$ must lie on the radical axis of $\Gamma$ and $\Gamma_B$. However, $P$ also lies on the radical axis of $\Gamma$ and $\Gamma_B$, since $\text{Pow}_{\Gamma}(P) = PC^2 = PA \cdot PB = \text{Pow}_{\Gamma_B}(P)$ (note that the $C$-Apollonius circle of $\triangle ABC$ swaps point $A$ and point $B$ under inversion and thus $PC^2 = PA \cdot PB$) which implies that line $\overline{PQ}$ is the radical axis of circles $\Gamma$ and $\Gamma_B$, implying that point $Q$ lies on a fixed-line. However, we see that since point $C$ and point $J$ are the centers of the circles $\Gamma$ and $\Gamma_B$ respectively, it follows that line $\overline{PQ} \perp$ line $\overline{CJ}$. However we had already established that the angle bisector of $\angle BPC$ is perpendicular to line $\overline{CJ}$, which implies that point $Q$ lies on the angle bisector of $\angle BPC$, as desired.