Question

AB and CD are the smallest and the longest sides of a quadrilateral ABCD,prove that angle A is greater than angle C, angle B greater than angle 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 

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Hope this will help you......

Answer 2

Answer:

simmilarly you can do angle b is greater than angle b by joining bd