In the figure AB = CD, ∠OCD > ∠COD and ∠OAB < ∠AOB. Let’s prove that OB < OD

Given: a figure such that AB = CD, ∠OCD > ∠COD and ∠OAB < ∠AOB

To prove: OB < OD

Consider ΔABO,

Given ∠AOB > ∠OAB

From the figure, AB is opposite to ∠AOB, and OB is opposite to ∠OAB

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ AB > OB…………(i)

Now consider ΔCOD,

Given ∠OCD > ∠COD

From figure, CD is opposite to ∠COD, and OD is opposite to ∠OCD

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒OD > CD…………(ii)

And given AB = CD………..(iii)

Now substituting equation (iii) in equation (i), we get

⇒ CD > OB…………(iv)

Now comparing equation (ii) and equation (iv), we get

OD > OB

Or OB < OD

Hence Proved.