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$.
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$