Question

prove that parallelogram circumscribing a circle is a rhombus ​

Answer 1

Answer:Answer is in the attachment...

It was done by me earlier in my notebook...n sorry for the trigonometry part

Step-by-step explanation:

Answer 2

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 VC≈AD ________2

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

THEREFORE;

DR=DS,

AP=AS,

BP=BQ,

AND

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

HOPE IT'S HELP U

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