In a trapezoid ABCD with AB parallel to CD, the diagonals AC and BD intersect at E. What is the ratio of the area of △ABE to the area of △CDE?

Correct option is A

​Given :

In trapezoid ABCD, AB||CD.
Diagonals AC and BD intersect at E.

Formula Used :

Triangles formed between parallel sides and intersecting diagonals are similar.

For similar triangles:
​$$\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{\text{Corresponding side}_1}{\text{Corresponding side}_2}\right)^2$$​​

Solution :

Since AB$$\parallel C$$​D, triangles $$\triangle$$​ABE and $$\triangle$$​CDE are similar.

Corresponding bases are AB and CD.

Therefore,
​$$\frac{\text{Area of } \triangle ABE}{\text{Area of } \triangle CDE}$$​​

​$$= \left(\frac{AB}{CD}\right)^2$$​​