Question -
Answer -
In ΔDOC,
∠CDO + ∠COD + ∠DCO = 1800 [Anglesum property of a triangle]
or 1/2∠CDA + ∠COD + 1/2∠DCB = 1800
∠COD =1800 – 1/2(∠CDA + ∠DCB) …..(i)
Also
We know, sum of all angles of a quadrilateral = 3600
∠CDA + ∠DCB = 3600 –(∠DAB + ∠CBA) ……(ii)
Substituting (ii) in (i)
∠COD =1800 – 1/2{3600 – (∠DAB + ∠CBA) }
We can also write, ∠DAB = ∠A and ∠CBA = ∠B
∠COD =1800 − 1800 +1/2(∠A + ∠B))
∠COD =1/2(∠A + ∠B)
Hence Proved.