29. In the diagram below, $A B C$ is a triangle with $A B=39 \mathrm{~cm}, B C=45$ $\mathrm{cm}$ and $C A=42 \mathrm{~cm}$. The tangents at $A$ and $B$ to the circumcircle of $\triangle A B C$ meet at the point $P$. The point $D$ lies on $B C$ such that $P D$ is parallel to $A C$. It is given that the area of $\triangle A B D$ is $x \mathrm{~cm}^{2}$. Find the value of $x$.
29. Answer: 168
First $\angle B D P=\angle B C A=\angle B A P$ so that $P, B, D, A$ are concyclic. Thus $\angle A C D=$ $\angle P B A=\angle P D A=\angle D A C$ so that $D A=D C$.

By cosine rule, $\cos C=3 / 5$. Thus $D C=\frac{1}{2} A C / \cos C=21 \times 5 / 3=35$. Hence $B D=10$ and $B C=10+35=45$. Thus area $(\triangle A B D)=\frac{10}{45} \times$ area $(\triangle A B C)$. By Heron's formula, area $(\triangle A B C)=756$. Thus area $(\triangle A B D)=\frac{10}{45} \times 756=168$.