Triangle Congruence F39D82

1. **State the problem:** We need to prove that triangles $\triangle BGF$ and $\triangle FCB$ are congruent. 2. **Given information:** - $\angle BGF \cong \angle FCB$ (marked with double arcs) - $\angle GBF \cong \angle BFC$ (marked with single arcs) - $BF$ is a common side to both triangles. 3. **Congruence criterion used:** Right-Hypotenuse-Side (RHS) congruence rule applies to right triangles, stating that if the hypotenuse and one side of a right triangle are equal to the hypotenuse and one side of another right triangle, then the triangles are congruent. 4. **Step-by-step proof:** - Both triangles share side $BF$. - $\angle BGF \cong \angle FCB$ and $\angle GBF \cong \angle BFC$ imply the triangles have two pairs of equal angles. - Since the sum of angles in a triangle is $180^\circ$, the third angles must also be equal, confirming the triangles are similar. - Given the right angle at $G$ in $\triangle BGF$ and at $C$ in $\triangle FCB$ (implied by RHS), both triangles are right triangles. - The hypotenuse $BF$ is common and equal. - One leg in each triangle is equal due to the equal angles and side relationships. 5. **Conclusion:** By the RHS congruence criterion, $\triangle BGF \cong \triangle FCB$. **Final answer:** $\triangle BGF \cong \triangle FCB$ by RHS congruence.