Question

In quadrilateral ABCD, side DC is largest
Show that AB + AD > DC - BC.

Answer 1

Hence Proved.

Step-by-step explanation:

Given,

In quadrilateral $ABCD$ and side $DC$ is largest.

Then Prove: $AB+AD>DC-BC$

From figure,

We know,

In Any triangle, Sum of two sides is greater then third sides.

In $\triangle ABD$,

$AB+AD>BD$__1

Also know,

In any triangle, Different of two sides allows less then third sides.

In $\triangle BCD$,

$DC-BC<BD$__2

From equation-1 and 2,

$AB+AD>BD>DC-BC$

We can also write,

∴ $AB+AD>DC-BC$

Hence Proved.

Answer 2

Proved below.

Step-by-step explanation:

Given:

In quadrilateral ABCD, side DC is largest.

As shown in the figure, join BD.

To prove:

AB + AD > DC - BC.

Proof:

In ABD,

we know that, property of a triangle that in a triangle, sum of two sides is always grteater that the third side.

AB + AD > BD        [1]

In CDB,

we know that, property of a triangle that in a triangle, difference of two sides is less that third side.

DC - BC < BD        [2]

Form Eq (1) and (2), we get

AB + AD > BD > DC - BC

Therefore, AB + AD > DC - BC.

Hence proved.