Question

In a quadrilateral ABCD, the line segment bisecting ∠C and ∠D at E. Prove that angle ∠A + ∠B = 2 ∠CED.​

Answer 1

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Question:-

•In a quadrilateral ABCD, the line segment bisecting∠C and ∠D at E. Prove that angle ∠A + ∠B = ∠CED.

To prove:-

•∠A + ∠B = 2∠CED

Proof:-

Let CE and DE be the bisectors of ∠C and ∠D respectively.

Then , ∠1 = ½∠C and ∠2=½∠D

In △DEC, we have ∠1 + ∠2 + ∠CED = 180° (sum of the angles of a △ is 180°)

==> ∠CED = 180° – ( ∠1 + ∠2 ) ...(i)

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

∴ ∠A + ∠B + ∠C + ∠D = 360°

==> ½( ∠A + ∠B ) + ½∠C + ½∠D = 180°

==> ½( ∠A + ∠B ) + ∠1 + ∠2 = 180°

==> ½( ∠A + ∠B) = 180° – ( ∠1 + ∠2 ).

From (i) and (ii), we get ½( ∠A + ∠B ) = ∠CED

Hence, ∠A + ∠B = 2∠CED

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Answer 2

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