Solve

 

 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.

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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°

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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).

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Step 3: Find angles EBC and FCB

Since ∠ABC = 50°, its exterior angle ∠EBC = 180° - 50° = 130°

Similarly, ∠FCB = 180° - ∠ACB = 180° - 50° = 130°

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Step 4: Angle Bisectors

Since D is the intersection of angle bisectors of ∠EBC and ∠FCB:

∠EBD = ∠CBD = 130° ÷ 2 = 65°

∠DCF = ∠DCF = 130° ÷ 2 = 65°

Now triangle BDC has:

∠CBD = 65°

∠DCE = 65°

So:

\angle BDC = 180° - 65° - 65° = 50°

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Final Answer:

\boxed{\angle BDC = 50^\circ}

Let me know if you’d like this visualized in a diagram!