Question

In a $\triangle ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $BD = CE$. If $\angle B = \angle C$, then show that $DE \parallel BC$.

Answer 1

Step 1: Understand the given problem:
We are given a triangle \( \triangle ABC \), where \( D \) and \( E \) are points on the sides \( AB \) and \( AC \), respectively, such that \( BD = CE \). We are also told that \( \angle B = \angle C \), and we need to prove that \( DE \parallel BC \).

Step 2: Use the basic property of similar triangles:
Since \( \angle B = \angle C \) and \( BD = CE \), we have two pairs of corresponding angles that are equal: \( \angle B \) and \( \angle C \), and \( \angle DBA \) and \( \angle ECA \). These equal angles suggest that \( \triangle BDE \) and \( \triangle CED \) might be similar.

Step 3: Prove the similarity of triangles \( BDE \) and \( CED \):
Consider triangles \( BDE \) and \( CED \). We have: - \( \angle B = \angle C \) (given), - \( \angle DBA = \angle ECA \) (vertically opposite angles), and - \( BD = CE \) (given). Thus, by the AA (Angle-Angle) criterion for similarity, we can conclude that \( \triangle BDE \sim \triangle CED \).

Step 4: Use the property of similar triangles to conclude parallelism:
Since \( \triangle BDE \sim \triangle CED \), the corresponding sides of these triangles are proportional. That is: \[ \frac{BE}{DE} = \frac{DE}{EC} \] This implies: \[ BE \cdot EC = DE^2 \] Now, if the ratio of the corresponding sides is constant and proportional, this indicates that line \( DE \) divides the triangle in a way that \( DE \parallel BC \). Therefore, we conclude that \( DE \parallel BC \).

Conclusion:
We have shown that if \( BD = CE \) and \( \angle B = \angle C \), then \( DE \parallel BC \), as required.