I2.4 In the figure, $A P$ bisects $\angle B A C$. Given that $A B=c, B P=d$, $P C=75$ and $A C=150$, find $d$.
Let $\angle B A P=\theta=\angle C A P, \angle A P C=\alpha, \angle B P C=180^{\circ}-\alpha$ $\frac{d}{\sin \theta}=\frac{100}{\sin \left(180^{\circ}-\alpha\right)} \cdots$
(1) and $\frac{75}{\sin \theta}=\frac{150}{\sin \alpha} \ldots$
$(1) \div(2) \Rightarrow d=50$