2003 USAMO Problems/Problem 4

Problem

Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$ , respectively. Lines $AB$ and $DE$ intersect at $F$ , while lines $BD$ and $CF$ intersect at $M$ . Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$ .

Solution

by April

Take $G\in BD: \,FG\parallel CD$ . We have:

$MF = MC\Longleftrightarrow \textrm{the quadrilateral}\; CDFG\; \textrm{is a parallelogram} \\ \Longleftrightarrow FD\parallel CG\Longleftrightarrow\angle FDA = \angle GCD\Longleftrightarrow\angle FDA + \angle CGF = 180^\circ \\ \Longleftrightarrow \angle ABE + \angle CGF = 180^\circ\Longleftrightarrow\textrm{the quadrilateral}\;CBGF\;\textrm{is cyclic} \\ \Longleftrightarrow\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM \\ \Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD = MC^{2}$

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2003 USAMO Problems/Problem 4

Problem

Solution

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