Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Saudi Arabia

Quadrilateral $ABCD$ with perpendicular diagonals $AC$ and $BD$ is inscribed in a circle. Altitude $DE$ in triangle $ABD$ intersects diagonal $AC$ in $F$ . Prove that $FB = BC$ .

Solution

$F$ is the orthocenter of $\triangle ABD$ , that is $BF \perp AD$ . We have $\widehat{BCA} \equiv \widehat{BDA}$ ( $ABCD$ is cyclic) and $\widehat{BDA} = \frac{\pi}{2} - \widehat{PAD}$ . Also, $\widehat{BFP} = \widehat{AFB'} = \frac{\pi}{2} - \widehat{FAB'} = \frac{\pi}{2} - \widehat{PAD}$ . It follows, $\widehat{BCA} = \widehat{BFC}$ , i.e. $FB = BC$ .

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