ABCD is a quadrilateral in which the bisectors of ∠A and ∠C meet DC produced at Y and BA produced at X respectively. Prove that .
Answers:
∠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)