23. In triangle $A B C, A B=28, B C=21$ and $C A=14$. Points $D$ and $E$ are on $A B$ with $A D=7$ and $\angle A C D=\angle B C E$. Find the length of $B E$.
23. Ans: 12
Let $B E=x$. By cosine rule,
$$
\cos B=\frac{21^{2}+28^{2}-14^{2}}{2(21)(28)}=\frac{7}{8} \text { and } \cos A=\frac{14^{2}+28^{2}-21^{2}}{2(14)(28)}=\frac{11}{16} .
$$

Thus $C D^{2}=14^{2}+7^{2}-2(14)(7)\left(\frac{11}{16}\right)=\frac{441}{4}$ so that $C D=\frac{21}{2}$.
Also
$$
\cos \angle A C D=\frac{14^{2}+\left(\frac{21}{2}\right)^{2}-7^{2}}{2(14)\left(\frac{21}{2}\right)}=\frac{7}{8} \text {. }
$$

Thus $\angle B=\angle A C D=\angle B C E$ so that triaugle $B E C$ is isosceles. Therefore.
$$
x=\frac{B C}{2 \cos B}=\frac{21}{2} \frac{8}{7}=12 .
$$