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 properties of the trapezium:
We are given a trapezium $ABCD$ with $AB \parallel DC$ and diagonals $AC$ and $BD$ intersecting at point $E$. We are also given that $ \triangle AED \sim \triangle BEC$. We need to prove that $AD = BC$.

Step 2: Using the property of similar triangles:
Since $ \triangle AED \sim \triangle BEC$, their corresponding sides are proportional. This gives us the following proportion:
\[ \frac{AE}{BE} = \frac{AD}{BC} = \frac{ED}{EC} \] From the above, we focus on the second part of the proportion, which gives us:
\[ \frac{AD}{BC} = \frac{AE}{BE} \]

Step 3: Using the parallel sides property:
Since $AB \parallel DC$, we know that the triangles formed by the diagonals and the parallel sides are similar, and the ratios of the corresponding sides are preserved.

Step 4: Conclusion:
Since the triangles $AED$ and $BEC$ are similar and we have established that $\frac{AD}{BC} = \frac{AE}{BE}$, the proportion tells us that $AD = BC$. Therefore, we have proven that $AD = BC$.