Concept:In an isosceles triangle, angles opposite equal sides are equal. Angle bisectors from the equal base angles are also equal in length.
Explanation:Given BA = BC, so ∠BAC = ∠BCA (base angles). AE bisects ∠A, so ∠BAE = ∠EAC. CF bisects ∠C, so ∠BCF = ∠FCA. Thus, ∠EAC = ∠FCA.
Now compare ΔAFC and ΔCEA:
1. AC is a common side.
2. ∠FAC = ∠ECA (half of equal base angles).
3. Because the triangle is isosceles, the bisectors AE and CF are equal in length. Hence, AF = CE? Wait careful: The existing solution says AE = CF. But ΔAFC and ΔCEA: side AF vs CE? Actually, the given is: ΔAFC ≅ ΔCEA. The sides involved: AFC has sides AF, FC, AC; CEA has sides CE, EA, AC. To use SAS, we need two sides and included angle. Common side AC. Angle at C in AFC is ∠FCA; angle at A in CEA is ∠EAC, which are equal. Then we need AF = CE? Or FC = EA? The original says AE = CF, but that would correspond to FC = EA. Indeed, in ΔAFC, side FC is opposite ∠FAC; in ΔCEA, side EA is opposite ∠ECA. So by SAS with AC common, ∠FCA = ∠EAC, and FC = EA. So that works: Because AE and CF are equal, and they correspond as FC (in AFC) and EA (in CEA). So we can state: Since AE = CF, we have FC = EA.
Thus, by SAS congruence (AC common, ∠FCA = ∠EAC, FC = EA), we get ΔAFC ≅ ΔCEA.
Answer:Option C: Δ AFC ≅ Δ CEA
