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