Question

IN a quadrilateral ABCD the line segment bisecting angle c and angle d meet at E prove that angle A + angle B=2 angle CED

Answer 1

hence,we can conclude that <A+<B=2<CED

Answer 2

SOLUTIONS:

Let CE and DE be the bisectors of ∠C and ∠D respectively. Then, ∠1 = 1/2∠C and ∠2 = 1/2∠D

In △DEC, we have

→ ∠1 + ∠2 + ∠CED = 180° [Sum of angles of a triangle]

→ ∠CED = 180° - (∠1 + ∠2). .....(i)

Again, the sum od the angles of a quadrilateral is 360°.

∴ ∠A + ∠B + ∠C + ∠D = 360°

→1/2(∠A + ∠B) + 1/2∠C + 1/2∠D = 180°

→1/2(∠A + ∠B) + ∠1+ ∠2 = 180°

→1/2(∠A + ∠B) = 180° - (∠1+ ∠2)....(ii)

From (i) and (ii), we get 1/2(∠A + ∠B) = ∠CED

Hence, ∠A + ∠B = 2∠CED