# 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

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by Ravikesh

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by Ravikesh

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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