Question

AB and cd are the smallest and the longest sides of a equal ABCD. Show that angle B is greater then angle D​

Answer 1

Solution:

Given:

In quadrilateral ABCD,

AB is the smallest side.         ....(1)

CD is the longest sides.        ....(2)

To Prove:

$\angle B>\angle D$

Construction:

Join B and D.

Mark the angles as shown in the figure.

Proof:

In △ABD,

$AD > AB$        (From(1))

$\therefore \angle 5 > \angle 6$        $\boxed{\text{(Angle opposite to the longer side is greater)}}$

In △CBD,

$CD > BC$        (From(2))

$\therefore \angle 7 > \angle 8$        $\boxed{\text{(Angle opposite to the longer side is greater)}}$

Adding equations

$\angle 5 + \angle 7 > \angle 6 + \angle 8\\\implies \angle B > \angle D$

Hence Proved.