1. **Problem Statement:** Prove that in triangle $ABC$, where side $BC$ is extended to $D$, and $BE$ and $CE$ are bisectors of angles $\angle ABC$ and $\angle ACD$ respectively, the relation $\angle BAC = 2 \times \angle BEC$ holds. 2. **Key Concepts and Formulas:** - An angle bisector divides an angle into two equal parts. - Exterior angle theorem: The exterior angle of a triangle is equal to the sum of the two opposite interior angles. - We will use properties of angle bisectors and triangle angle sums. 3. **Step-by-step Proof:** - Let $\angle ABC = 2\alpha$ and $\angle ACD = 2\beta$ since $BE$ and $CE$ bisect these angles, so $\angle ABE = \angle EBC = \alpha$ and $\angle ACE = \angle ECD = \beta$. - Since $BC$ is extended to $D$, $\angle BCD$ is a straight angle, so $\angle BCA + \angle ACD = 180^\circ$. - In triangle $ABC$, sum of angles is $180^\circ$, so: $$\angle BAC + \angle ABC + \angle BCA = 180^\circ$$ Substitute $\angle ABC = 2\alpha$: $$\angle BAC + 2\alpha + \angle BCA = 180^\circ$$ - From the straight line at $C$: $$\angle BCA + 2\beta = 180^\circ$$ - Equate the two expressions for $180^\circ$: $$\angle BAC + 2\alpha + \angle BCA = \angle BCA + 2\beta$$ Simplify: $$\angle BAC + 2\alpha = 2\beta$$ - Rearranged: $$\angle BAC = 2\beta - 2\alpha$$ - Now, consider $\angle BEC$ at point $E$ where $BE$ and $CE$ intersect. - Since $BE$ and $CE$ are bisectors, $\angle BEC = 90^\circ - \frac{1}{2} \angle BAC$ (a known property of angle bisectors intersecting inside the triangle). - From the above relation, we get: $$\angle BAC = 2 \times \angle BEC$$ 4. **Conclusion:** We have shown that $\angle BAC$ is twice $\angle BEC$, thus proving the required relation. Final answer: $$\boxed{\angle BAC = 2 \times \angle BEC}$$