Problem
GEO-B1-M01-P016 A Point on a Side and Equal Segments
#16
★★★☆☆ Level 3 of 5
In triangle \(ABC\), \(\angle A=60^\circ\), \(\angle B=50^\circ\). Point \(D\) lies on side \(AB\), and \(CD=BD\). Find \(\angle ACD\).
Find \(\angle C\), then use isosceles triangle \(BCD\).
In triangle \(ABC\), \(\angle C=180^\circ-60^\circ-50^\circ=70^\circ\). Since \(D\) lies on \(AB\), angle \(CBD\) equals \(\angle CBA=50^\circ\). In triangle \(BCD\), sides \(CD\) and \(BD\) are equal, so the base angles at \(BC\) are equal: \(\angle BCD=\angle CBD=50^\circ\). Hence \(\angle ACD=\angle ACB-\angle BCD=70^\circ-50^\circ=20^\circ\).
A good example where equal segments must be translated into equal angles.