AoPS - Problem

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Problem

Source: China Western Mathematical Olympiad 2002 P2

Tags: geometry, parallelogram, circumcircle, geometry proposed

Fang-jh

27.12.2008 14:07

Let $O$ be the circumcenter of acute triangle $ABC$ . Point $P$ is in the interior of triangle $AOB$ . Let $D,E,F$ be the projections of $P$ on the sides $BC,CA,AB$ , respectively. Prove that the parallelogram consisting of $FE$ and $FD$ as its adjacent sides lies inside triangle $ABC$ .

SaYaT

28.12.2008 15:55

Fang-jh wrote: Let $O$ be the circumcenter of acute triangle $ABC$ . Point $P$ is in the interior of triangle $AOB$ . Let $D,E,F$ be the projections of $P$ on the sides $BC,CA,AB$ , respectively. Prove that the parallelogram consisting of $FE$ and $FD$ as its adjacent sides lies inside triangle $ABC$ .

Let

$\angle BAC=\alpha$

$\angle ABC=\beta$

$\angle ACB=\gamma$ Let

$P$ is forth point of parallelogram

$FEKD$ . It's enough to prove that

$\angle AEF+\angle FEK<180^o$ and

$\angle BDF+\angle FDK<180^o$ Let's prove it: We have:

$90^o+\angle OBP>90^o$

$<=>$

$\gamma+\angle PAF+\angle OAP+\angle OBP>90^o$ So

$\angle EAP+\angle PBD=\gamma+\angle OAP+\angle OBP>90^o-\angle PAF=\angle APF=\angle AEF$

$<=>$

$\angle EFD>\angle AEF$

$<=>$

$\angle AEF+180^o-\angle EFD<180^o$

$<=>$

$\angle AEF+\angle FEK<180^o$ Similiarly we achieve that

$\angle BDF+\angle FDK<180^o$ So parallelogram

$FEKD$ lies inside

$\triangle ABC$