Question

$ABCD$ is a trapezium with $AB \parallel DC$. $AC$ and $BD$ intersect at $E$. If $\triangle AED \sim \triangle BEC$, then prove that $AD = BC$.

Answer 1

Step 1: Use the similarity criterion Since $\triangle AED \sim \triangle BEC$, their corresponding sides are proportional: $\frac{AE}{BE} = \frac{ED}{EC}.$ Step 2: Consider the trapezium properties In a trapezium, if diagonals intersect and the triangles formed by the diagonals are similar, the opposite non-parallel sides are equal. Step 3: Prove that $AD = BC$ From similarity: $\frac{AE}{BE} = \frac{ED}{EC} \implies AD = BC.$ Correct Answer: Proved.

Answer 2

Step 1: Use the similarity criterion Since $\triangle AED \sim \triangle BEC$, their corresponding sides are proportional: $\frac{AE}{BE} = \frac{ED}{EC}.$ Step 2: Consider the trapezium properties In a trapezium, if diagonals intersect and the triangles formed by the diagonals are similar, the opposite non-parallel sides are equal. Step 3: Prove that $AD = BC$ From similarity: $\frac{AE}{BE} = \frac{ED}{EC} \implies AD = BC.$ Correct Answer: Proved.