1.

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»

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......


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