Question

In the following figure, FD || BC || AE and AC || ED. Find the value of x.

Answer 1

Given: FD || BC || AE and AC || ED.

Construction: Produce DF such that it intersect AB at G.

In triangle ABC,

∠A + ∠B + ∠C = 180°  ...[Angle sum property of triangle]

52° + 64° + ∠C = 180°

∠C = 180° – (52° + 64°)

∠C = 180° – 116°

∠C = 64°

Now, as see that DG || BC and DG || AE,

∠ACB = ∠AFG   ...[FG || BC and FC is a transversal. So, corresponding angles]

64° = ∠AFG

Also, GFD is a straight line.

So, ∠GFA + ∠AFD = 180°  ...[Linear pair]

64° + ∠AFD = 180°

∠AFD = 180° – 64°

∠AFD = 116°

Also, FD || AE and AF || ED

Hence, AEDF is a parallelogram.

Now, ∠AFD = ∠AEF   ...[Opposite angles in a parallelogram are equal]

∠AED = x = 116°