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

Since $\triangle ABD \sim \triangle ACE$ (Angle-Angle Similarity, $\angle B = \angle C$), we have:
$\frac{BD}{AB} = \frac{CE}{AC}$
Given $BD = CE$, it follows that:
$\frac{BD}{AB} = \frac{CE}{AC} = \frac{1}{k}, \quad k = proportionality constant.$
Thus, $DE \parallel BC$ (by Basic Proportionality Theorem).
Correct Answer: Proved

Answer 2

Since $\triangle ABD \sim \triangle ACE$ (Angle-Angle Similarity, $\angle B = \angle C$), we have:
$\frac{BD}{AB} = \frac{CE}{AC}$
Given $BD = CE$, it follows that:
$\frac{BD}{AB} = \frac{CE}{AC} = \frac{1}{k}, \quad k = proportionality constant.$
Thus, $DE \parallel BC$ (by Basic Proportionality Theorem).
Correct Answer: Proved