1. **Problem statement:** Prove that $\hat{A}BD = \hat{A}CE$ using the given figure with triangle $ABC$ and points $D$ and $E$ on the line through $B$ and $C$. 2. **Key idea:** Angles subtended by the same segment or corresponding angles formed by parallel lines are equal. 3. **Step 1:** Note that points $B$, $C$, $D$, and $E$ lie on the same straight line. 4. **Step 2:** Since $D$ is on the left of $B$ and $E$ is on the right of $C$, the line $DE$ extends beyond $BC$. 5. **Step 3:** Consider triangles $ABD$ and $ACE$. Since $AB$ and $AC$ are sides of triangle $ABC$, and $D$ and $E$ lie on the line through $B$ and $C$, angles $\hat{A}BD$ and $\hat{A}CE$ are alternate interior angles formed by transversal $AC$ or $AB$ with line $DE$. 6. **Step 4:** By the Alternate Interior Angles Theorem, these angles are equal: $$\hat{A}BD = \hat{A}CE$$ 7. **Conclusion:** We have shown that $\hat{A}BD = \hat{A}CE$ as required.