Question

How can we prove that "If a circle touches all the four sides of a quadrilateral ABCD at points PQRS. Then AB+CD = BC + DA"

Answer 1

HELLO DEAR,

we can observe from the figure ;

DR = DS (Tangents on the circle from point D)--------( 1 )

AP = AS (Tangents on the circle from point A)---------( 2 )

BP = BQ (Tangents on the circle from point B)---------( 3 )

CR = CQ (Tangents on the circle from point C)---------( 4 )

now, adding all these equations,----( 1 ) , -----( 2 ) , ------( 3 ) & -----( 4 )

we get,

DR + AP + BP + CR = DS + AS + BQ + CQ

⇒ (BP + AP) + (DR + CR)  = (DS + AS) + (CQ + BQ)

⇒ CD + AB = AD + BC

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Let us consider a quadrilateral ABCD, and a circle is circumscribed by ABCD
Also, Sides AB, BC, CD and DA touch circle at P, Q, R and S respectively.
To Proof : AB + CD = AD + BC
Proof :
In the Figure,
As tangents drawn from an external point are equal.
We have
AP = AS [tangents from point A]
BP = BQ [tangents from point B]
CR = CQ [tangents from point C]
DR = DS [tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence Proved.