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 conditions:
We are given a triangle $ABC$ with points $D$ and $E$ on sides $AB$ and $AC$ respectively. The following conditions are provided:
- $BD = CE$
- $\angle B = \angle C$
We need to prove that $DE \parallel BC$.

Step 2: Use the Angle-Angle (AA) Similarity Criterion:
Since $\angle B = \angle C$, triangles $BDE$ and $CDE$ share two corresponding angles, namely $\angle BDE = \angle CDE$ (as these are alternate interior angles formed by the transversal $DE$). Therefore, by the Angle-Angle (AA) similarity criterion, we can conclude that triangles $BDE$ and $CDE$ are similar.

Step 3: Apply the property of similar triangles:
Since triangles $BDE$ and $CDE$ are similar, the corresponding sides of these triangles are proportional. This gives us the following proportionality:
\[ \frac{BD}{DE} = \frac{CE}{DE} \] But from the given condition, we know that $BD = CE$. Therefore, we can substitute this into the proportion:
\[ \frac{BD}{DE} = \frac{BD}{DE} \] This shows that the ratio of corresponding sides is equal, confirming that the lines $DE$ and $BC$ are parallel by the Basic Proportionality Theorem (also known as Thales' Theorem).

Conclusion:
We have shown that $DE \parallel BC$, as required.