China Western Mathematical Olympiad

Proof As shown in the figure, we construct a parallelogram $DEFG$ with $FE$ and $FD$ as adjacent sides. To prove the proposition to be true, we need only to prove that $\angle FEG<\angle FEC$, and $\angle FDG<\angle FDC$. It is equivalent to proving: $\angle BFD<\angle BAC$, and $\angle AFE<\angle ABC$.

In fact, we construct $OH$ with $OH\perp BC$, and $H$ is the foot of the perpendicular. From $PD\perp BC$ and $PF\perp AB$, we know that four points $B$, $F$, $P$ and $D$ are concyclic. Thus $\angle BFD=\angle BPD$. But $\angle PBD>\angle OBH$, hence $90^{\circ }-\angle PBD<90^{\circ }-\angle OBH$, and that is, $\angle BPD<\angle BOH$. Moreover, $O$ is the circumcenter of $△ABC$, so $\angle BOH=\frac{1}{2}\angle BOC=\angle BAC$. Therefore, $\angle BFD=\angle BPD<\angle BOH=\angle BAC$, that is, $\angle BFD<\angle BAC$.

Similarly, we can prove that $\angle AFE<\angle ABC$. Therefore, the proposition holds.