Problem
GEO-B1-M05-P028 Angle Between Altitudes
#28
★★★★★ Level 5 of 5
In acute triangle \(ABC\), altitudes \(BD\) and \(CE\) meet at point \(H\). Prove that \(\angle DHE=180^\circ-\angle A\).
Remember that \(HD\) lies on the altitude from \(B\), and \(HE\) lies on the altitude from \(C\).
Since \(BD\perp AC\), line \(HD\) is perpendicular to \(AC\). Since \(CE\perp AB\), line \(HE\) is perpendicular to \(AB\). The angle between two lines equals the angle between their perpendiculars or supplements it to \(180^\circ\), depending on the chosen angle. Inside triangle \(DHE\), the angle is obtuse, so \(\angle DHE=180^\circ-\angle BAC=180^\circ-\angle A\).
Although the solution is short, the problem is often difficult: one must see the angle between altitudes as the angle between perpendiculars to sides.