Perpendicular Line Geometry

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Perpendicular Line Geometry

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Source: KJMO 2005

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#1 h May 20, 2018, 1:21 PM • 2 Y

Y by Adventure10, Mango247

In $\triangle ABC$ , let the bisector of $\angle BAC$ hit the circumcircle at $M$ . Let $P$ be the intersection of $CM$ and $AB$ . Denote by $(V,WX,YZ)$ the intersection of the line passing $V$ perpendicular to $WX$ with the line $YZ$ . Prove that the points $(P,AM,AC), (P,AC,AM), (P,BC,MB)$ are collinear.

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#2 h May 20, 2018, 7:24 PM • 2 Y

Y by Adventure10, Mango247

Let $(P,AM,AC)=N, (P,AC,AM)=L, (P,BC,MB)=K$ ,then $L$ is the orthocenter of $APN$ so $LNP=\frac{A}{2}$ .on the other hand,since $PKM=90-A=MPK$ so $MK=MP$ ;also $M$ lies on the perpendicular bisector of $PN$ ,so:
$MP=MK=MN \implies \text{M is the center of (PKN)} \implies KNP=\frac{A}{2} \implies \text{N,L,K are collinear}$

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