Problem
GEO-B1-M01-P013 Altitude and Angle Bisector
In triangle \(ABC\), altitude \(AH\) and angle bisector \(AL\) are drawn from vertex \(A\). Prove that the angle between \(AH\) and \(AL\) equals half the difference of angles \(B\) and \(C\).
Express \(\angle BAL\) through \(\angle B\) and \(\angle C\), and the angle between \(AB\) and \(AH\) through \(\angle B\).
Let \(\angle B=\beta\), \(\angle C=\gamma\), and assume \(\beta>\gamma\). Then \(\angle A=180^\circ-\beta-\gamma\), so \(\angle BAL=\frac{180^\circ-\beta-\gamma}{2}\). Since \(AH\perp BC\), the angle between \(AB\) and \(AH\) is \(90^\circ-\beta\). Thus the angle between \(AH\) and \(AL\) equals \(\frac{180^\circ-\beta-\gamma}{2}-(90^\circ-\beta)=\frac{\beta-\gamma}{2}\). If \(\gamma>\beta\), the same argument gives \(\frac{\gamma-\beta}{2}\).
This is the first problem in the module where it is useful to discuss two cases or the absolute difference.