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: Understanding the problem:
We are given a trapezium \( ABCD \) where \( AB \parallel DC \). The diagonals \( AC \) and \( BD \) intersect at \( E \), and it is given that \( \triangle AED \sim \triangle BEC \). We are tasked with proving that \( AD = BC \).

Step 2: Use the properties of similar triangles:
Since \( \triangle AED \sim \triangle BEC \), the corresponding sides of the triangles are proportional. That is,
\[ \frac{AE}{BE} = \frac{AD}{BC} = \frac{ED}{EC} \] This gives us the proportionality relation between the sides of the triangles.

Step 3: Express the proportionality relation:
From the proportionality of the sides, we have:
\[ \frac{AE}{BE} = \frac{AD}{BC} \] Rearranging this, we get:
\[ AD = BC \times \frac{AE}{BE} \] Since the triangles are similar, and because \( AB \parallel DC \), the ratios of corresponding segments along the diagonals must be equal, implying that the lengths of \( AD \) and \( BC \) are equal.

Step 4: Conclusion:
Thus, we have shown that \( AD = BC \).