Question

In a quadilateral ABCD the line segments bisecting angle C and angle D meet at E.Prove that angle A+angle B=2 angle DEC

Answer 1

ABCD is a quadrilateral. CE and DE are the bisector of ∠C and ∠D respectively.
 CE is the bisector of ∠C.
 DE is the bisector of ∠D.
 In ΔCED, ∠DCE + ∠CDE + ∠CED = 180° (Angle sum property)
In quadrilateral ABCD,
 ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of quadrilateral is 360°)
 ∴ ∠A + ∠B + 360° – 2∠CED = 360° [Using (3)]
⇒ ∠A + ∠B = 2 ∠CED