In a triangle ABC, ∠BCA = 50°. 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 ∠FDE, in degrees is equal to
Explanation:
In the figure above:
In ∆ABC, ∠A + ∠B + ∠C = 180° ⇒ x + y = 130°
Since DA = DE ⇒ ∠DAE = ∠DEA = x ⇒ ∠ADE = 180° – 2x
Also, DB = DF ⇒ ∠DBF = ∠DFB = y ⇒ ∠BDC = 180° – 2y
∠ADE + ∠EDF + ∠FDB = 180° ⇒ 180 – 2x + ∠EDF + 180 – 2y = 180 ⇒ ∠EDF = 2x + 2y – 180 ⇒ ∠EDF = 260° – 180° = 80°
Hence, option (c).
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