angle bisector wanted, <PAB = <PCB =1/4 (<A+ < C)

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angle bisector wanted, <PAB = <PCB =1/4 (<A+ < C)

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Source: Ukrainian Geometry Olympiad 2020, X p4

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#1 h Apr 27, 2020, 1:57 AM

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Inside triangle $ABC$ , the point $P$ is chosen such that $\angle PAB = \angle PCB =\frac14 (\angle A+ \angle C)$ . Let $BL$ be the bisector of $\vartriangle ABC$ . Line $PL$ intersects the circumcircle of $\vartriangle APC$ at point $Q$ . Prove that the line $QB$ is the bisector of $\angle AQC$ .

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#2 h Apr 27, 2020, 9:35 AM • 1 Y

Y by TurtleKing123

Let $M=BL\cap(ABC)$ ,

1- With angle working you get that $M$ is the center of $(APC)$ .

2- Let $K$ be the midpint of arc $APC$ in $(APC)$ . With angle working you get $\angle PMB=\angle BMK$ , then $K$ is the reflection of $P$ through $BM$ .

3- Using power of point, $PL\cdot LQ=AL\cdot LC=BL\cdot LM$ , then $BPMQ$ is cyclic, and $PM=MQ$ implies $\angle PBM=\angle MBQ$ , then $B,K, Q$ are collinear and we are done.

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#3 h Apr 27, 2020, 11:22 AM

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hectorraul wrote:

Let $M=BL\cap(ABC)$ ,

1- With angle working you get that $M$ is the center of $(APC)$ .

2- Let $K$ be the midpint of arc $APC$ in $(APC)$ . With angle working you get $\angle PMB=\angle BMK$ , then $K$ is the reflection of $P$ through $BM$ .

3- Using power of point, $PL\cdot LQ=AL\cdot LC=BL\cdot LM$ , then $BPMQ$ is cyclic, and $PM=MQ$ implies $\angle PBM=\angle MBQ$ , then $B,K, Q$ are collinear and we are done.

Similar solution. A nicer approach than mine I would say.

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#4 h Apr 27, 2020, 12:23 PM • 3 Y

Y by Mango247, Mango247, Mango247

Tuymaada 2012 #3

This post has been edited 1 time. Last edited by Kamran011, Apr 27, 2020, 12:24 PM

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