Mathematics - Properties of Triangle Question with Solution | TestHub

Let $BE$ and $CF$ be the altitudes from vertices B and C respectively, intersecting at the orthocentre H. In the quadrilateral $AEHF$, we have $\angle AEH=90^{\circ }$ and $\angle AFH=90^{\circ }$. The sum of angles in a quadrilateral is $360^{\circ }$, so $\angle FHE+\angle A+\angle AEH+\angle AFH=360^{\circ }$. This means $\angle FHE+A+90^{\circ }+90^{\circ }=360^{\circ }$, which simplifies to $\angle FHE=180^{\circ }-A$. Since $\angle BHC$ and $\angle FHE$ are vertically opposite angles, $\angle BHC=\angle FHE=180^{\circ }-A$.