Math, asked by riyabhargaV11, 9 months ago

A circle touches all side of a quadrilateral ABCD prove that AB+VD=BC+DA.​

Answers

Answered by CᴀɴᴅʏCʀᴜsʜ
1

Answer:

The circle touches on four points on the quadrilateral at P, Q, R, and S.

From the diagram, we can conclude that tangent from exterior point of circle are equal. Hence

AP = AQ

DQ = DR

RC = SC

PB = BS

Adding all the above equations, we get

AP + DQ + RC + PB = AQ + DR + SC + BS

AQ + DQ + SC +BS =AP  + PB + DR + RC ...... ( i )

From the figure given below, we get,

AQ+DQ = AD

SC+BS = BC

AP+PB = AB

DR+RC=CD

Substituting the above values in equation (i)

AD + BC = AB + CD

Hence proved.

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