Question
Prove that parallelogram circumscribing a circle is a rhombus.
✓
Answer
Given ABCD is a parallelogram in which all the sides touch a given circle
To prove:- ABCD is a rhombus
Proof:-
$\because$ ABCD is a parallelogram
$\therefore$ AB = DC and AD = BC
Again AP, AQ are tangents to the circle from the point A
$\therefore$ AP = AQ
Similarly, BR = BQ
CR = CS
DP = DS
$\therefore$(AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS)
$\Rightarrow$ AD + BC = AB + DC
$\Rightarrow$ BC + BC = AB + AB [$\because$ AB = DC, AD = BC]
$\Rightarrow$ 2BC = 2AB
$\Rightarrow$ BC = AB
Hence, parallelogram ABCD is a rhombus
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