Question

PA and PB are tangents to the circle with centre O from an external point P, touching the circle at A and B respectively. Show that the quadrilateral AOBP is cyclic
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Answer 1

$\huge \tt{solution}$

Since tangents to a circle is perpendicular to the radius

∴OA is perpendicular AP and OB is perpendicular BP

⇒∠OAP=90

and ∠OBP=90

⇒∠OAP+∠OBP=90

+90

=180

......(1)

In quadrilateral OAPB,

∠OAP+∠APB+∠AOB+∠OBP=360

⇒(∠APB+∠AOB)+(∠OAP+∠OBP)=360

⇒∠APB+∠AOB+180

=360

⇒∠APB+∠AOB=360

−180

=180

......(2)

From (1) and (2), the quadrilateral AOBP is cyclic.

Answer 2

➸ Given :-

→ PA and PB are tangents to

the circle with centre o from an

external point P.

➸ To prove :-

→ Quadrilateral AOBP is cyclic

➸ Prove :-

We know that the tangent at any point of a circle is perpendicular to radius through the point of contact.

⟶ PA⊥OA, i.e., ∠OAP = 90° and ....(i)

⟶ PB ⊥ OB, i.e., ∠OBP = 90°. ....(ii)

Now, the sum of all the angles of a quadrilateral is 360°

∴ ∠AOB + ∠OAP + ∠APB + ∠OBP = 360°

⟶ ∠AOB+ ∠APB = 180° (using (i) and (ii)]

∴ quadrilateral OAPB is cyclic

[since both pairs of opposite angles have the sum 180°]

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Note :-

Refer the above attachment....