Question

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see adjacent figure). Show that ∠A >∠C and ∠B >∠D.

Answer 1

Solution:

Given:

In quadrilateral ABCD, AB smallest & CD is longest sides.

To Prove: ∠A>∠C

& ∠B>∠D

Construction: Join AC.

Mark the angles as shown in the figure..

Proof:

In △ABC , AB is the shortest side.

BC > AB

∠2>∠4 …(i)

[Angle opposite to longer side is greater]

In △ADC , CD is the longest side

CD > AD

∠1 >∠3 …(ii)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

∠2+∠1 >∠4+∠3

∠A >∠C

Similarly, by joining BD, we can prove that

∠B >∠D

Hope this will help you......