Problem
GEO-B1-M01-P014 Angle Between Two Bisectors
#14
★★★☆☆ Level 3 of 5
In triangle \(ABC\), the internal angle bisectors of \(B\) and \(C\) meet at \(I\). It is known that \(\angle BIC=124^\circ\). Find \(\angle A\).
In triangle \(BIC\), the angles at \(B\) and \(C\) are halves of angles \(B\) and \(C\).
In triangle \(BIC\), \(\angle IBC=\frac{B}{2}\) and \(\angle ICB=\frac{C}{2}\). Hence \(\angle BIC=180^\circ-\frac{B+C}{2}\). Since \(B+C=180^\circ-A\), we get \(\angle BIC=180^\circ-\frac{180^\circ-A}{2}=90^\circ+\frac{A}{2}\). Thus \(124^\circ=90^\circ+\frac{A}{2}\), so \(A=68^\circ\).
It is useful to derive \(\angle BIC=90^\circ+\frac{A}{2}\), not give it as a memorised formula.