22. The figure below shows a circle with diameter $A B . C$ and $D$ are points on the circle on the same side of $A B$ such that $B D$ bisects $\angle C B A$. The chords $A C$ and $B D$ intersect at $E$. It is given that $A E=169 \mathrm{~cm}$ and $E C=119 \mathrm{~cm}$, If $E D=x \mathrm{~cm}$, find the value of $x$.
22. Answer: 65
Since $B E$ bisects $\angle C B A$, we have $\frac{B C}{B A}=\frac{E C}{E A}=\frac{119}{169}$. Thus we can let $B C=119 y$ and $B A=169 y$ for some real number $y$. Since $\angle B C A=90^{\circ}$, we have
$$
\begin{array}{l}
A B^{2}=A C^{2}+B C^{2} \\
(169 y)^{2}=(169+119)^{2}+(119 y)^{2} \\
y^{2}(169-119)(169+119)=(169+119)^{2} \\
y^{2}=\frac{169+119}{169-119}=\frac{144}{25} \\
y=\frac{12}{5}
\end{array}
$$

Hence, from triangle $B C E$, we have $B E=\sqrt{119^{2}+(119 y)^{2}}=119 \times \frac{13}{5}$.
Finally, note that $\triangle A D E$ and $\triangle B C E$ are similar, so we have
$$
E D=\frac{A E \times C E}{B E}=\frac{169 \times 119}{119 \times \frac{13}{5}}=65 \mathrm{~cm} \text {. }
$$