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m∠FDE = 52°

Solution:

Given data:

DE ≅ DF, CD || BE, BC || FD and m∠ABF = 116°

Sum of the adjacent angles in a straight line = 180°

m∠ABF + m∠CBF = 180°

116° + m∠CBF = 180°

m∠CBF = 64°

If CD || BE, then CD || BF.

Hence CD || BE and BE || FD.

Therefore BFCD is a parallelogam.

In parallelogram, Adjacent angles form a linear pair.

m∠CBF + m∠BFD = 180°

64° + m∠BFD = 180°

m∠BFD = 116°

Sum of the adjacent angles in a straight line = 180°

m∠BFD + m∠DFE = 180°

116° + m∠DFE = 180°

m∠DFE = 64°

we know that DE ≅ DF.

In triangle, angles opposite to equal sides are equal.

m∠DFE = m∠DEF

m∠DEF = 64°

sum of all the angles of a triangle = 180°

m∠DFE + m∠DEF + m∠FDE = 180°

64° + 64° + m∠FDE = 180°

m∠FDE = 52°

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