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Home » From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that: i) ∠AOP = ∠BOP ii) OP is the ⊥ bisector of chord AB.
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that: i) ∠AOP = ∠BOP ii) OP is the ⊥ bisector of chord AB.
Class 10 | Exercise 18A | ICSE | Maths | Selina | Tangents and Intersecting Chords
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that: i) ∠AOP = ∠BOP ii) OP is the ⊥ bisector of chord AB.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(A) - 14

According to the given question,

  1. i) In 

        \[\vartriangle AOP\text{ }and\text{ }\vartriangle BOP\]

    , we have

    \[AP\text{ }=\text{ }BP\]

    \[\left[ Tangents\text{ }from\text{ }P\text{ }to\text{ }the\text{ }circle \right]\]

    \[OP\text{ }=\text{ }OP\left[ Common \right]\]

    \[OA\text{ }=\text{ }OB[Radii\text{ }of\text{ }the\text{ }same\text{ }circle]\]

By SAS criterion of congruence

    \[\Delta AOP\cong \Delta BOP\]

By C.P.C.T we have

    \[\angle AOP\text{ }=\angle BOP\]

  1. ii) In

        \[\vartriangle OAM\text{ }and\text{ }\vartriangle OBM\]

    , we have

    \[OA\text{ }=\text{ }OB\text{ }\left[ Radii\text{ }of\text{ }the\text{ }same\text{ }circle \right]\]

    \[\angle AOM\text{ }=\angle BOM\text{ }~\text{ }[Proved~\angle AOP\text{ }=\angle BOP]\]

    \[OM\text{ }=\text{ }OM\text{ }\left[ Common \right]\]

By SAS criterion of congruence

    \[\Delta OAM\cong \Delta OBM\]

By C.P.C.T we have

    \[AM\text{ }=\text{ }MB\]

And

    \[\angle OMA\text{ }=\angle OMB\]

Since,

    \[\angle OMA\text{ }+\angle OMB\text{ }=\text{ }{{180}^{o}}\]

So,

    \[\angle OMA\text{ }=\angle OMB\text{ }=\text{ }{{90}^{o}}\]

Therefore,

    \[OM\text{ }or\text{ }OP\]

is the perpendicular bisector of

    \[chord\text{ }AB.\]

– Hence Proved

i

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