Question

ABCD is a quadrilateral in which the bisectors of angle A and angle C meet DC prodiced at Y an BA produced at X respectively.
Prove that, angle X+ angle Y = 1/2 (angle A + angle C)

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Answer 1

Step-by-step explanation:

∠X + ∠Y = 1/2 (∠A + ∠C)

Here, ∠1 = ∠2 and ∠3 and ∠4

In △XBC , we have

∠X + ∠B + ∠4 = 180°

∠X + ∠B + 1/2 ∠C = 180°

In △ADY , we have

∠2 + ∠D +∠Y = 180°

1 / 2 + ∠A + ∠D + ∠Y = 180°

Adding (i) and (ii) , we have

∠X + ∠Y + ∠B + ∠D + 1 / 2 ∠C + 1 / 2 ∠A = 360°

Also, in quadrilateral ABCD,

∠A + ∠B + ∠C + ∠D = 360°

∴ ∠X + ∠Y + ∠B + ∠D + 1 / 2 ∠C + 1 / 2 ∠A = ∠A + ∠B + ∠C + ∠D

∠X + ∠Y = ∠A - 1 / 2 ∠A + ∠C - 1 / 2 ∠C

∠X + ∠Y = 1 / 2 ( ∠A + ∠C)