Question

If a circle touches all the four sides a quadrilateral ABCD at the points PQRS. Then show that

AB + CD = BC + DA.​

Answer 1

To Prove :

AB + CD = BC + DA.

Given :

For Quadrilateral,

• We have Sides AB,BC,CD and DA.

For Circle,

• and, We have a circle touch all sides of Quadrilateral.

Solution :

Taking, Circle.

Sides of Quadrilateral touch to circle Let sides be,

• P , Q , R and S respectively.

Now,

• BP , BQ , QC , CR , DR, DS, SD , SA and AP are tangent, respect to Circle.

So We can say that, by Circle theorem

• AP = AS.
• BP = BQ.
• CR = CQ.
• DR = DS.

Taking, Quadrilateral

• It have property, Opposite sides are equal.

We have,

$\tt AP = AS. - - (1) \\ \tt BP = BQ. - - (2)\\ \tt CR = CQ. - - (3) \\ \tt DR = DS . - - (4)$

Adding Equation 1,2,3,4, we get.

$\dashrightarrow \tt AP + BP + CR + DR= AS +BQ + CQ + DS .$

$\dashrightarrow \tt (AP + BP) + (CR +DR) = (AS + DS)+ (BQ + CQ).$

$\dashrightarrow \tt AB + CD= BC + AD.$

Hance Proved.

$\rule{300}3$

Some information :

Properties regarding Circle,

• If two tangent drawn to a circle from external point, then tangent are equal in length.

Answer 2

Step-by-step explanation:

Given : ABCD is a quadrilateral .

To Prove : AB+CD=BC+DA

Solution :

Tangents from  the same point touches to circle are equal

Then ,

AP = AS [ tangent from point " A " ]

BP = BQ [ tangent from point " B " ]

DR = DS [ tangent from point " D " ]

CR = CQ [ tangent from point " C " ]

Add equations

AP + BP + DR + CR = AS + BQ + SD + QC

(AP + PB) + (DR + RC) = (AS + SD) + (BQ + QC)

Hence Proved

Thanks