Triangle Congruence 19Cc0E

1. **State the problem:** We are given an isosceles triangle $\triangle ABC$ with base $BC$, and perpendiculars $BE$ and $CF$ dropped from vertices $B$ and $C$ onto sides $AC$ and $AB$ respectively. We need to prove that $BE \cong CF$. 2. **Recall properties and formulas:** - In an isosceles triangle, the two sides opposite the equal angles are equal. Here, since $BC$ is the base, $AB = AC$. - $BE \perp AC$ means $BE$ is perpendicular to $AC$. - $CF \perp AB$ means $CF$ is perpendicular to $AB$. - To prove segments congruent, we can use triangle congruence criteria such as SAS, ASA, or RHS. 3. **Consider triangles $\triangle ABE$ and $\triangle ACF$:** - $AB = AC$ (given, isosceles triangle) - $\angle BEA = \angle CFA = 90^\circ$ (given perpendiculars) - $AE$ is common to both triangles (segment on $AC$ and $AB$ respectively) 4. **Show $AE = AF$:** Since $E$ lies on $AC$ and $F$ lies on $AB$, and $AB = AC$, the segments $AE$ and $AF$ correspond to the same relative positions on equal sides. Because $BE$ and $CF$ are altitudes, $E$ and $F$ are feet of perpendiculars from $B$ and $C$ respectively. 5. **Use triangle congruence:** Triangles $\triangle ABE$ and $\triangle ACF$ have: - $AB = AC$ (isosceles triangle sides) - $\angle BEA = \angle CFA = 90^\circ$ (right angles) - $AE = AF$ (corresponding segments on equal sides) Therefore, by the RHS (Right angle-Hypotenuse-Side) congruence criterion, $\triangle ABE \cong \triangle ACF$. 6. **Conclude:** By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), $BE \cong CF$. **Final answer:** $$BE \cong CF$$