Question

prove that the tangents drawn from an external point of a circle are equal​

Answer 1

*********** ⭕Answer⭕***********

Step-by-step explanation:

Here in the above pik given a circle with centre O ,a point P lying outside the circle and two tangents PQ and PR on the circle from P . We are required to prove PQ = PR

For, this we join OP, OQ, and OR .Then angle OQP and angle ORP are right angle , because tangents are always perpendicular to the radius of a circle .

Now in ∆PQO and ∆POR

$\leadsto \bold{\red{ \angle{ \: Q \: } = \angle R \: \: \: }} \\ \leadsto \: \red{OQ \: = OR \: \: \: \: } \\ \leadsto \red{\: OP = O P \: \: \: \: \: \: \: \: \: }$

$\bold \blue {\therefore \: \triangle are \: congruent}$

$\huge \leadsto \red{\huge \: Hence \: QP = PR \: \: \: \: }$

Answer 2

$\huge\underline\mathbb {SOLUTION:-}$

Construction:

• Draw a circle centered at O.

Figure provided in the attachment.

Let PR and QR are tangent drawn from an external point R to the circle touching at points P and Q respectively.

• Join OR

Proof:

In ∆OPR and ∆OQR,

OP = OQ (Radii of the same circle)

$\angle$OPR = $\angle$OQR (Side PR and QR are tangents to the circle)

OR = OR (Common Side)

∆OPR $\cong$∆OQR (By R.H.S)

$\underline \mathsf \blue {Therefore \: :-}$

PR = QR (C.P.C.T)

• Thus, tangent drawn from an external point to a circle are equal.

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