Question

In a triangle ABC, $\angle BCA = 50^\circ$. D and E are points on AB and AC, respectively, such that $AD = DE$. If F is a point on BC such that $BD = DF$, then $\angle FDE$, in degrees, is equal to

Accepted Answer

B. Let $\angle A = \alpha$ and $\angle B = \beta$. In $	riangle ABC$, $\alpha + \beta + 50^\circ = 180^\circ \implies \alpha + \beta = 130^\circ$. In $	riangle ADE$, since $AD = DE$, it is an isosceles triangle with $\angle AED = \angle DAE = \alpha$. Thus, $\angle ADE = 180^\circ - 2\alpha$. In $	riangle BDF$, since $BD = DF$, it is an isosceles triangle with $\angle DFB = \angle DBF = \beta$. Thus, $\angle BDF = 180^\circ - 2\beta$. Since points A, D, B lie on a straight line: $$\angle BDF + \angle FDE + \angle ADE = 180^\circ$$ $$(180^\circ - 2\beta) + \angle FDE + (180^\circ - 2\alpha) = 180^\circ$$ $$\angle FDE = 2(\alpha + \beta) - 180^\circ = 2(130^\circ) - 180^\circ = 260^\circ - 180^\circ = 80^\circ$$.

Angle in a triangle with equal segments

Explanation

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