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(a) Prove that the lengths of two tangents drawn from an external point to a circle are equal.
OR
(b) If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
OR
(b) If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
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(a)
Given: $PQ$ and $PR$ are tangents to a circle with centre $O$
To prove: $PQ = PR$.
Construction: Join $OP, OQ$ and $OR$.
Proof :In $\Delta OQP$ and $\Delta ORP$
$\angle OQP = \angle ORP = 90^\circ$
$OQ = OR$ (Radii)
$OP = OP$ (Common side)
$\implies \Delta OQP \cong \Delta ORP$ [By RHS congruence criterion]
$PQ = PR$ [CPCT]
OR
(b)
As lengths of tangents drawn from an external point to a circle are equal
$AP = AS \dots (i)$
$BP = BQ \dots (ii)$
$DR = DS \dots (iii)$
$CR = CQ \dots (iv)$
Adding the above equations we get, $AB + CD = AD + BC$
$\implies 2AB = 2 AD$ [$\because$ Opp. Sides of a Parallelogram are equal]
$\implies AB = AD \implies \parallel gm ABCD$ is a rhombus.
Given: $PQ$ and $PR$ are tangents to a circle with centre $O$
To prove: $PQ = PR$.
Construction: Join $OP, OQ$ and $OR$.
Proof :In $\Delta OQP$ and $\Delta ORP$
$\angle OQP = \angle ORP = 90^\circ$
$OQ = OR$ (Radii)
$OP = OP$ (Common side)
$\implies \Delta OQP \cong \Delta ORP$ [By RHS congruence criterion]
$PQ = PR$ [CPCT]
OR
(b)
As lengths of tangents drawn from an external point to a circle are equal
$AP = AS \dots (i)$
$BP = BQ \dots (ii)$
$DR = DS \dots (iii)$
$CR = CQ \dots (iv)$
Adding the above equations we get, $AB + CD = AD + BC$
$\implies 2AB = 2 AD$ [$\because$ Opp. Sides of a Parallelogram are equal]
$\implies AB = AD \implies \parallel gm ABCD$ is a rhombus.