Problem
GEO-B2-M01-P012 The Angle at the Orthocenter
#12
★★★☆☆ Level 3 of 5
In acute triangle \(ABC\), the altitudes meet at \(H\). Prove that \(\angle BHC=180^\circ-\angle BAC\).
Use \(BH\perp AC\) and \(CH\perp AB\).
Since \(BH\perp AC\), the angle between \(BH\) and \(AB\) equals \(90^\circ-\angle A\). Since \(CH\perp AB\), the angle between \(CH\) and \(AC\) is also expressed through \(90^\circ\). Summing around the vertex gives \(\angle BHC=180^\circ-\angle BAC\). In other words, the angle between two altitudes is supplementary to the angle between the corresponding sides.
This is a standard fact often used in problems with circles and altitudes.