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: Analyze given information
Consider triangle $ABC$ with points $D$ on side $AB$ and $E$ on side $AC$. The provided conditions are:
- $BD = CE$
- $\angle B = \angle C$
The objective is to demonstrate that $DE \parallel BC$.

Step 2: Employ the Angle-Angle (AA) Similarity Criterion
Given $\angle B = \angle C$, triangles $BDE$ and $CDE$ share two equal corresponding angles. Specifically, $\angle BDE = \angle CDE$ because they are alternate interior angles formed by the transversal $DE$. Therefore, by the AA similarity criterion, triangles $BDE$ and $CDE$ are similar.

Step 3: Apply similarity properties
As triangles $BDE$ and $CDE$ are similar, their corresponding sides are in proportion:
\[\frac{BD}{DE} = \frac{CE}{DE}\]Since $BD = CE$ (given), we substitute this into the proportion:
\[\frac{BD}{DE} = \frac{BD}{DE}\]This equality confirms that the ratio of corresponding sides is equal. Consequently, by the Basic Proportionality Theorem (Thales' Theorem), lines $DE$ and $BC$ are parallel.

Conclusion:
It has been proven that $DE \parallel BC$.