Question

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

Answer 1

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In a quadrilateral ABCD the line segment bisecting ∠C and ∠D meet at E.prove that ∠A + ∠B = 2∠CED

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

$then \: \angle \: 1 = \frac{1}{2} \angle \: c \: and \: \angle \: 2 = \frac{1}{2} \angle \: d.$

In ∆DEC,we have ∠1 + ∠2 + ∠CED = 180°

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(Sum of a ∆ is 180°)

⇒∠CED = 180° - (∠1 + ∠2).

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

$\therefore \: \angle \: A \: + \angle \: B + \angle \: C + \angle \: D = 360\degree$

$⇒ \frac{1}{2} (\angle \: A + \angle \: B) + \frac{1}{2} \angle \: C + \frac{1}{2} \angle \: D = 180\degree$

$⇒ \frac{1}{2} (\angle \: A + \angle \: B) + \angle \: 1 + \angle \: 2 = 180\degree$

$⇒ \frac{1}{2} (\angle \: A + \angle \: B) = 180\degree - (\angle \: 1 + \angle \: 2).$

$from \: (i) \: and \: (ii) \: we \: get \: \frac{1}{2} (\angle \: A + \angle \: B) = \angle \: CED$

Hence,∠A + ∠B = 2∠CED.

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