Let $\angle BAE = x$ and $\angle FCB = y$ $\frac{sin \angle BAE}{sin \angle CAE} = \frac{sin x}{sin (A-x)}$ In $\triangle ABM$, $\frac{sin x}{sin (A-x)} = \frac{AM}{AB}.\frac{BE}{EM}$ In $\triangle BPM$, $\frac{BE}{EM} = \frac{BP}{PM}.\frac{sin \angle BPE}{sin \angle MPE}$ But $\angle BPE = 90-A$, $\angle MPE = 90$ and $PM=AM$ Therefore, $\frac{BE}{EM} = \frac{BP.cos A}{AM}$ Hence, $\frac{sin x}{sin (A-x)}=\frac{BP.cos A}{AB}=cos A.cos B$ So, $\frac{sin \angle BAE}{sin \angle CAE} = cos A.cos B$ Similarly, $\frac{sin y}{sin (C-y)}=cos B.cos C$ So, $\frac{sin \angle ACF}{sin \angle BCF}=\frac{1}{cos B.cos C}$ $\frac{sin \angle CBH}{sin \angle ABH}=\frac{cos C}{cos A}$ Therefore, $\frac{sin \angle BAE}{sin \angle CAE}.\frac{sin \angle ACF}{sin \angle BCF}.\frac{sin \angle CBH}{sin \angle ABH}=cos A.cos B.\frac{1}{cos B.cos C}.\frac{cos C}{cos A} = 1$ Hence, by converse of Trigonometric form of Ceva's theorem, $AE, CF$ and $BH$ are concurrent.