Two tangents `P Aa n dP B` are drawn from an external point `P` to a circle with centre `Odot` Prove that `A O B P` is a cyclic quadrilateral.

Explanation

PA and PB are tangents to the circle

`:. OA _|_ PA and OB_|_ PB `(tangent to the circle is perpendicular to radius at the of contact)

`:. /_A and /_B are 90^0`

In quadrilateral `AOBP`

`/_A + /_O +/_B+/_P = 360^`
0
`:. 90^0 + /_O + 90^0 + /_P =360`
^0
`:. /_O + /_P = 180^0`

Sum of opposite angles of quadrilateral `AOBP is 180^0`

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