In a convex quadrilateral $ABCD$ we have $AB\cdot CD=BC\cdot DA$ and $2\angle A+\angle C=180^\circ$. Point $P$ lies on the circumcircle of triangle $ABD$ and is the midpoint of the arc $BD$ not containing $A$. It is known that the point $P$ lies inside the quadrilateral $ABCD$. Prove that $\angle BCA=\angle DCP$ Proposed by S. Berlov
Problem
Source: tuymaada 2003
Tags: geometry, circumcircle, angle bisector, geometry unsolved
pontios
25.08.2006 18:45
Let $F \in BD \cap AP$. Then $AF$ is the angle bisector of $\angle BAD \Rightarrow \frac{FD}{FB}= \frac{AD}{AB}\Rightarrow$ $\frac{FD}{FB}= \frac{CD}{CB}\Rightarrow CF$ is the angle bisector of $\angle DCB$ Let $w=(O,R)$ be the circumcircle of $\triangle ABD$ $2A+C=180 \Rightarrow A<90$ So $\angle BOD = 2A$ as a result, $C$ is on the circumcircle $k$ of $\triangle ODB$ (on the arc $BD$ which doesn't contain $O$)
Quote:
$A$ is a point on the exterior of $w=(O,R)$ and $M$ is the point of intersection of $w$ and the ray $OA$.
Let $B,C \in w$ and $D \in OA \cap BC$. Then:
$O,A,B,C$ lie on the same circle $\Leftrightarrow D$ is inverse of $A$ w.r.t. $w\Leftrightarrow M$ is the incenter of $\triangle ABC$
http://www.mathlinks.ro/Forum/viewtopic.php?t=108097
From the lemma we get that the points $F,C$ are inverse points w.r.t. $w$ Again according to the lemma, since $F \in (AP) \Rightarrow$ the angle bisector of $\angle ACP$ is the ray $CO$ Let $\phi = \angle OCA = \angle OCP$ and $\theta = \angle OCD = \angle OCB$ $\angle BCA = \angle BCO+\angle OCA = \theta+\phi$ $\angle DCP = \angle DCO+\angle OCP = \theta+\phi$
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j555
25.08.2006 21:51
I think that my solution is simpler: Let $X$ be a point lying in exterior of $ABCD$ s.t. $CX=CD$, $\angle BCX=\angle A$. We have: $\triangle DCX \sim \triangle DPB$ and $\triangle BCX \sim \triangle BAD$, so $\triangle DCP \sim \triangle DXB \sim \triangle ACB$.Thus $\angle BCA=\angle DCP$.