Trapezium 1

We note that $\triangle ABE$ and $\triangle CDE$ are similar. Let $AE=a$ and $\dfrac {CC}{AB} = \dfrac {EC}{AE} = k$ . Then $CD = 45k$ and $EC = ak$ . Since $AC = AE + EC$ , $\implies a + ak = 100$ and $a = \dfrac {100}{1+k}$ .

Also $\triangle ABE$ and $\triangle ACD$ are similar. Therefore

$\begin{aligned} \frac {DC}{AE} & = \frac {AC}{AB} \\ \frac {45k}a & = \frac {100}{45} \\ 81 k & = 4\blue a = \frac {400}{1+k} & \small \blue{\text{Note that }a = \frac {100}{1+k}} \\ 81 k^2 + 81 k & = 400 \\ \left(k + \frac 12\right)^2 & = \frac {400}{81}+\frac 14 \\ \implies k & = \sqrt{\frac {1681}{324}} - \frac 12 = \frac {16}9 \\ \implies CD & = 45 k = 45 \times \frac {16}9 = \boxed{80} \end{aligned}$