Question

In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 1/2 (∠A + ∠B).​

Answer 1

Answer:

In ∆COD

<COD + <1 + <2 = 180°

<COD = 180° - (<1 +<2)

<COD = 180° - 1/2(<C+<D)

<COD = 180°- 1/2 [360° -(<A+<B)]

<COD = 1/2 (<A+<B)

.......... please like and follow guy's ️