Question

If all the stides of a parallogram youch a circle prove that it is a rhombus

Answer 1

Answer:

your answer is bro

Step-by-step explanation:

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Answer 2

$\huge\underline\mathfrak{Correct\:Question-}$

If all the sides of a parallelogram touch a circle then, prove that it is a rhombus.

$\huge\underline\mathfrak{Solution-}$

Given :

• ABCD is a parallelogram and touches a circle of centre O.

To prove :

• ABCD is a rhombus.

Proof :

We know that, tangents from an external point are equal.

$\implies$ AP = AS ________(1)

$\implies$ BP = BQ ________(2)

$\implies$ CR = CQ ________(3)

$\implies$ DR = DS ________(4)

By adding (1), (2), (3) and (4)

$\implies$ AP + BP + CR + DR = AS + BQ + CQ + DS

$\implies$ ( AP + BP ) + ( CR + DR ) = ( AS + DS ) + ( BQ + CQ )

$\implies$ AB + CD = AD + BD _____(A)

Also, ABCD is a parallelogram, therefore, opposite sides are equal.

So, AB = CD and BC = AD ______(B)

Put the value of (B) in (A).

AB + AB = AD + AD

$\implies$ 2AB = 2AD

$\implies$ $\cancel{2}$AB = $\cancel{2}$ AD

$\implies$ AB = AD

Since, the adjacent sides of a parallelogram are also equal.

Therefore, it is a rhombus.

Hence proved!