Question

In a quadrilateral ABCD, the line segments
bisecting angleC and angkleD meet at E. Prove that
angleA+ angleB = 2angleCED.
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Answer 1

Answer:

this is your solution for the following information

Answer 2

Step-by-step explanation:

Given:

• In quadrilateral ABCD line segments bisecting ∠C and ∠D meet at E. (see fig)

To Prove:

• ∠A + ∠B = 2∠CED

Proof: 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 the ∠s of a triangle is 180°)

=> ∠CED = 180° – (∠1 + ∠2).............(1)

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

$∴ ∠A + ∠B + ∠C + ∠D = 360° \: \\ \\ \implies \: \frac{1}{2} (∠A + ∠B) + \frac{1}{2} ∠C + \frac{1}{2} ∠D \: = 180° \\ \\ \implies \frac{1}{2} (∠A + ∠B) + ∠1 + ∠2 = 180° \: \\ \\ \implies \: \frac{1}{2} (∠A + ∠B) = 180° - (∠1 + ∠2)......(2)$

From (1) and (2) , we get

$\frac{1}{2} (∠A + ∠B) = ∠CED \\ \\ Hence, ∠A + ∠B = 2∠CED \:$