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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.8 AIME Prove it Saudi Arabia

In an acute triangle $ABC$ the angle bisector $AL$ , $L \in BC$ , intersects its circumcircle at $N$ . Let $K$ and $M$ be the projections of $L$ onto sides $AB$ and $AC$ . Prove that triangle $ABC$ and quadrilateral $AKNM$ have equal areas.

Solution

Let $P$ be the second point of intersection of segment $BC$ and the circle circumscribed about quadrilateral $AKLM$ . Denote by $E$ the intersection point of the lines $KN$ and $BC$ and by $F$ the intersection point of the lines $MN$ and $BC$ .

Then $\widehat{BCN} = \widehat{BAN}$ and $\widehat{MAL} = \widehat{MPL}$ , as angles on the same arc. Since $AL$ is a bisector, $\widehat{BCN} = \widehat{BAL} = \widehat{MAL} = \widehat{MPL}$ , and consequently $PM \parallel NC$ . Similarly we prove $KP \parallel BN$ . Then the quadrilaterals $BKPN$ and $NPMC$ are trapezoids; hence
$S_{BKE} = S_{NPE} \quad \text{and} \quad S_{NPF} = S_{CMF}.$
Therefore $S_{ABC} = S_{AKNM}$ .

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