Question

if the circle touches all sides of parallelogram internally then show that parallelogram is a rhombus​

Answer 1

Answer:

A circle touches all side of a parallelogram must be a rhombus.

Given:-

AB=CD

BC=AD

Now, AP=AS

If from one external point, two tangents are drawn two circle then have equal tangent segments.

So, AP=AS [tangent from point A] ------(1)

⇒BP=BQ [tangent from point B] ---------(2)

⇒CR=CQ [tangent from point C] ---------(3)

⇒DR=DS [tangent from point D] ----------(4)

equation (1)+(2)+(3)+(4)

AP+BP+CR+DR=AS+BQ+CQ+DS

AB+DC=AD+BC [from diagram]

2AB=2AD from ⊛

AB=AD And from ⊛ AB=CD and BC=AD

So,

AB=BC=CD=AD

It is rhombus.

Answer 2

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A circle touches all side of a parallelogram must be a rhombus.

So, option B is correct.

Given - AB=CD

BC=AD

Now, AP=AS

If from one external point, two tangents are drawn two circle then have equal tangent segments.

So, AP=AS [tangent from point A] ------(1)

⇒BP=BQ [tangent from point B] ---------(2)

⇒CR=CQ [tangent from point C] ---------(3)

⇒DR=DS [tangent from point D] ----------(4)

equation (1)+(2)+(3)+(4)

AP+BP+CR+DR=AS+BQ+CQ+DS

AB+DC=AD+BC [from diagram]

2AB=2AD from ⊛

AB=AD And from ⊛ AB=CD and BC=AD

So,

AB=BC=CD=AD

It is rhombus.