Problem
GEO-B1-M08-P016 A Circle on Altitudes
#16
★★★★☆ Level 4 of 5
In acute triangle \(ABC\), altitudes \(BD\) and \(CE\) meet at point \(H\). Prove that points \(A,D,H,E\) lie on one circle, and find \(\angle DHE\) in terms of \(\angle A\).
Find two right angles, then look at the angle between the altitudes.
Since \(BD\perp AC\), \(\angle ADH=90^\circ\). Since \(CE\perp AB\), \(\angle AEH=90^\circ\). Hence \(A,D,H,E\) lie on the circle with diameter \(AH\). Also, \(HD\perp AC\) and \(HE\perp AB\), so the obtuse angle between \(HD\) and \(HE\) equals \(180^\circ-\angle A\). Therefore \(\angle DHE=180^\circ-\angle A\).
A strong mixed problem: altitudes, a circle, and an angle between perpendiculars.