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Let's work through it step-by-step. We are given: Triangle ABC is isosceles (with AB = AC, since angle BAC is given). AE and AF are straight lines. ∠EBD = 65° ∠BAC = 80° D is the intersection of the bisectors of ∠EBC and ∠FCB. We are to find ∠BDC. --- Step 1: Triangle ABC Since triangle ABC is isosceles with ∠BAC = 80°, and the equal sides are AB = AC, the base angles ∠ABC and ∠ACB are equal. So, we calculate: \angle ABC = \angle ACB = \frac{180° - 80°}{2} = 50° --- Step 2: Using ∠EBD = 65° From the diagram description, EBD is part of angle EBC. So: ∠EBC = ∠EBD + ∠CBD We're told ∠EBD = 65° D lies on the bisector of ∠EBC, so ∠EBD = ∠CBD Thus, both are 65°, meaning: \angle EBC = \angle EBD + \angle CBD = 65° + 65° = 130° But earlier we found: \angle ABC = 50° That suggests EBC is an exterior angle at B, formed when AE extends past AB (same with FCB being an exterior angle at C). --- Step 3: Find angles EBC and FCB Since ∠ABC = 50°, its exterior angle ∠EBC = 180° - 50° = 130° ...
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