Geometry: parallel lines, similar triangles, show this=that

Question

I have trouble specifically with part (ii) I can see BC bisecting AD and therefore AE because BC || DE and because ABC and ADE are similar triangles (I already have a proof on that - AAA).

I just don't know how to put my arguments together...

Please help

Answer 1

Here's just one way to do it:

For $AC=CE$, the ratio $\frac{AC}{AE}=\frac{1}{2}$. If we can show this statement, we are done.

I assume you have proved that $\triangle ABC$ and $\triangle ABD$ are similar. Let $AB = x$ and hence $AD=2x$.

Then by similar triangles, it follows that

\begin{align} \frac{AC}{AE} & = \frac{AB}{AD} \\ & = \frac{x}{2x} \\ & = \frac{1}{2} \end{align}

We have shown that the ratio $\frac{AC}{AE}=\frac{1}{2} \implies AC=CE.$