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: Problem Statement:
Given a trapezium \( ABCD \) with \( AB \parallel DC \). Diagonals \( AC \) and \( BD \) intersect at \( E \). If \( \triangle AED \sim \triangle BEC \), prove that \( AD = BC \).

Step 2: Apply Similarity Properties:
Since \( \triangle AED \sim \triangle BEC \), their corresponding sides are proportional:
\[\frac{AE}{BE} = \frac{AD}{BC} = \frac{ED}{EC}\]
This establishes the proportionality of the sides.

Step 3: Derive the Relationship:
From the proportionality, we have:
\[\frac{AE}{BE} = \frac{AD}{BC}\]
Rearranging yields:
\[AD = BC \times \frac{AE}{BE}\]
Given that the triangles are similar and \( AB \parallel DC \), the ratios of corresponding segments of the diagonals are equal, which implies \( AD = BC \).

Step 4: Final Proof:
Therefore, it is proven that \( AD = BC \).