Question

Prove that the parallelogram circumscribing a
circle is a rhombus?
​

Answer 1

Step-by-step explanation:

SOLUTION:

GIVEN: ABCD IS A PARALLELOGRAM CIRCUMSCRIBING A CIRCLE

TO PROVE: AB= BC=CD≈DA

PROOF: ABCD IS ONE PARALLELOGRAM

THEREFORE AB ≈CD. 1

THEREFORE BC≈AD. 2

WE KNOW THAT THE TANGENTS DRAWN FROM AN EXTERNAL POINTS TO THE CIRCLES ARE EQUAL

THEREFORE;

DR=DS, AP≈AS,BP=BQ & CR=CQ

ADDING ALL THESE,WE GET

DR+CR+BP+AP≈DS+CQ+BQ+AS

»(BP+AP)+(DR+CR)≈(DS+AS)+(CQ+BQ)

»AB+CD≈AD+BC. 3

SUBSTITUTING 1 & 2 IN 3

2 AB =2 BC »AB = BC. 4,

FROM EQUATION 1,2 & 4,

AB ≈ BC ≈ CD ≈ DA

THEREFORE ABCD IS A RHOMBUS

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