Question

A circle touches all the four sides of the quadrilateral ABCD, prove that AB+CD=BC+DA​

Answer 1

Professional strategy :

To concept points are :-

• Tangent to a circle touches it at a single point only .
• The length of tangent drawn from an external point to a circle are equal .

Specific steps :

Given :

• $\color{red} {Circle\:touches\:all the\:four\:sides\:of\:the\: quadrilateral}$

That is the Circle is inscribed inside ABCD.

To Proove :

• AB + CD = BC + DA

That is AS, AP, BP, BQ, CQ, CR, DR, DS are tangents to the Circle from external points A,B,C,D

Let,

$\color{green}{The\: circle\:touches\: sides\: AB,\: BC,\:CD\:and\:DA\: at \:P,\: Q,\:R\: and\: S\: respectively.}$

Refer to the attachment above ..

From the figure AB, BC, CD and DA are tangents to the circle with point of P ,Q ,R and S respectively .

Therefore,

$\sf\huge\implies$ AP = AS ....(i)

$\sf\huge\implies$ BP = BQ ...(ii)

$\sf\huge\implies$ CR = CQ. ...(iii)

Now ,

$\sf\huge\implies$ L.H.S = AB + CD

$\sf\huge\implies$ ( AP + BP ) + ( PQ +DR) $\color{green} {From\:the\:figure}$

$\sf\huge\implies$ ( AS + BQ )+ ( CQ + DS ) $\color{green} {From\:(i),\:(ii),\:(iii),\:(iv)}$

$\sf\huge\implies$ ( BQ + CQ) + (AS +DS)

$\sf\huge\implies$ BC + DA

$\sf\huge\implies$ R.H.S

Hence proved,

$\sf\huge\implies$AB + CD = BC + DA

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