3ucm and the parallelogramonthe0adestandsparallelogram.24. AB and CD a

1.	3ucm and the parallelogramonthe0adestandsparallelogram.24. AB and CD are respectively the smallest and longest sidof quadrilateral ABCD. Show that22
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

3ucm and the parallelogramonthe0adestandsparallelogram.24. AB and CD are respectively the smallest and longest sidof quadrilateral ABCD. Show that22

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