17. In $\triangle A B C, A C>A B$, the internal angle bisector of $\angle A$ meets $B C$ at $D$, and $E$ is the foot of the perpendicular from $B$ onto $A D$. Suppose $A B=5, B E=4$ and $A E=3$. Find the value of the expression $\left(\frac{A C+A B}{A C-A B}\right) E D$.
17. Ans: 3
Let $D E=x, A C=a>5$ and $C D=u$. Then $B D=\sqrt{4^{2}+x^{2}}$ and $\cos \angle C A D=$ $\cos \angle B A D=3 / 5$. By the cosine rule applied to $\triangle A D C$, we have $u^{2}=(3+x)^{2}+$ $a^{2}-2 a(3+x)(3 / 5)$. Using the angle bisector theorem, we have $\frac{u^{2}}{4^{2}+x^{2}}=\frac{a^{2}}{5^{2}}$. Thus $25\left[(3+x)^{2}+a^{2}-2 a(3+x)(3 / 5)\right]=a^{2}\left(16+x^{2}\right)$.

This can be simplified to $\left(a^{2}-25\right) x^{2}+30(a-5) x-9(a-5)^{2}=0$. Since $a>5$, we can cancel a common factor $(a-5)$ to get $(a+5) x^{2}+30 x-9(a-5)=0$, or equivalently $(x+3)((a+5) x-3(a-5))=0$. Thus $x=3(a-5) /(a+5)$. From this, we obtain $\left(\frac{A C+A B}{A C-A B}\right) E D=\left(\frac{a+5}{a-5}\right) x=3$.