Q. 6

# In the figure AB

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

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