The diagram shows two circles that touch at C. A, B and C are points on the small circle, centre P. C, D and E are points on the large circle, centre Q. APCQE and BCD are straight lines.

1. Identify one pair of equal corresponding angles at C.

Since $AC$ and $EC$ are on the same straight line, and $BC$ and $DC$ are on the same straight line,

$\angle ACB = \angle ECD$

because they are vertically opposite angles.

2. Show a second pair of corresponding angles is equal using radii and straight lines.

In the small circle, $PB=PC$ because they are radii. Hence triangle $PBC$ is isosceles, so

$\angle PBC=\angle BCP.$

Since $A,P,C$ are collinear and $B,C,D$ are collinear,

$\angle ABC=\angle PBC \quad\text{and}\quad \angle BCP=\angle BCA.$

In the large circle, $QC=QD$ because they are radii. Hence triangle $QCD$ is isosceles, so

$\angle QCD=\angle CDQ.$

Since $C,Q,E$ are collinear and $B,C,D$ are collinear,

$\angle EDC=\angle CDQ.$

Also $PCQ$ is a straight line, so the angle that line $BCD$ makes with $PC$ equals the angle it makes with $CQ$ on the opposite side, giving

$\angle BCP=\angle QCD.$

Therefore

$\angle ABC=\angle EDC.$

Since two pairs of corresponding angles are equal,

$\triangle ABC \sim \triangle EDC$

by AA similarity.

1. Find the ratio of corresponding side lengths of the similar triangles.

Since $A, P, C$ are collinear and $PA=PC=5$,
$AC=5+5=10.$
Since $C, Q, E$ are collinear and $QC=QE=8$,
$CE=8+8=16.$
For similar triangles $ABC$ and $EDC$,
$AC:CE=10:16=5:8.$

2. Square the linear ratio to obtain the area ratio and express it in the form $1:n$.

For similar figures,
$\text{area ratio}=(\text{length ratio})^2.$
So
$\text{area of }\triangle ABC : \text{area of }\triangle EDC=5^2:8^2=25:64.$
In the form $1:n$,
$25:64=1:\frac{64}{25}=1:2.56.$