CAT 2023 Slot 1 · Quant · Question 18

Angles standing on the same arc of a circle are equal, and that is what makes the four small triangles at E pair off into similar ones.

Compare \triangle ABE and \triangle DCE: the angles at A and D both stand on arc BC, and the angles at B and C both stand on arc AD.

So \triangle ABE\sim \triangle DCE, with the ratio of similarity \frac{AB}{DC}=\frac{2}{1}.

Corresponding sides give \frac{BE}{CE}=\frac{AB}{DC}=\frac{2}{1}.

Now compare \triangle AED and \triangle BEC: the angles at D and C both stand on arc AB, and the angles at A and B both stand on arc DC.

So \triangle AED\sim \triangle BEC, with ratio \frac{AD}{BC}=\frac{4}{5}.

Corresponding sides give \frac{AE}{BE}=\frac{AD}{BC}=\frac{4}{5}.

Chaining the two results, \frac{AE}{CE}=\frac{AE}{BE}\times \frac{BE}{CE}.

Substituting, \frac{AE}{CE}=\frac{4}{5}\times \frac{2}{1}=\frac{8}{5}.

So AE:CE=8:5 — the third option.

Shortcut. The two similar pairs hand over \frac{AE}{BE} and \frac{BE}{CE}, and BE cancels when they are multiplied. So the answer is just the product of the two given ratios, \frac{4}{5}\times \frac{2}{1}, with the second one taken the right way round.