Question

A circle touches all sides of a parallelogram. So the parallelogram must be a
Give a suitable explanation please ​

Answer 1

Answer:

Given:  

A circle touches the quadrilateral ABCD in all four sides

To prove:

AB +CD = AD + BC

Solution:

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

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.  

Step-by-step explanation:

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