In the figure AB < OB and CD > OD let’s prove that ∠BAO > ∠OCD.

Given: a figure such that AB < OB and CD > OD

To prove: ∠BAO > ∠OCD

Consider ΔABO,

Given AB < OB

From 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.

⇒∠AOB < ∠OAB…………(i)

Now consider ΔCOD,

Given CD > OD

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.

⇒∠COD > ∠OCD…………(ii)

From figure we can see that

∠AOB = ∠COD……..(iii) (as they form vertically opposite angles)

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

⇒∠COD < ∠OAB…………(iv)

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

∠OAB > ∠OCD

Or ∠BAO > ∠OCD

Hence Proved