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Explanation:

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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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