Question

In quadrilateral abcd, do and co are the bisectors of angle d and angle c respectively. Pt angle of cod=1/2[a+b]

Answer 1

Step-by-step explanation:

Gn:

ABCD is a quadrilateral

CO and DO are bisectors of ∠C and ∠D respectively.

To prove:

∠COD = 1/2 (∠A + ∠B)

Proof:

Let ∠A = A, ∠B = B, ∠C = C, ∠D = D

taking ABCD:

A + B + C + D = 360 (∠ sum property)

A + B = 360 - (A+ B)

1/2 (C + D) = 1/2 (360 - (A + B))

1/2 (C + D) = 180 - 1/2 (A + B)

1/2 (C + D) - 180 = - 1/2 (A + B)

180 - 1/2 (C + D ) = 1/2 (A + B) [Multiplying -1 to the eq. ]

COD = 1/2 (A + B)

[ as COD is a triangle, so 1/2 (C + D) + COD = 180 ]

Hence Proved!

Answer 2

Answer:

We have to prove that COD = 1/2 (a + b )

Solution :

$\qquad \sf \: COD = \angle \: 1 + \angle \: 2 = 180 \degree \\ \\ \\ \implies \sf \angle COD = 180 - ( \angle1 + \angle2) \\ \\ \\ \implies \sf \angle \: COD = 180 - ( \frac{1}{2} \: \angle \: C + \frac{1}{2} \: \angle \: D) \\ \\ \\ \implies \sf \angle \: COD = 180 - \frac{1}{2} \: ( \: \angle \: C + \angle \: D) \\ \\ \\ \implies \sf \angle \: COD = 180 - \frac{1}{2} \: [360 - ( \: \angle \: a + \: \angle \: b)] \\ \\ \\ \implies \sf \angle \orange {COD = \frac{1}{2} \: ( \: \angle \: a + \: \angle \: b)}$