Question

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary.

Thenks! ^•^​

Answer 1

ANSWER :

Given      : PA And PB are two tangents drawn from external

                  point P.

To Prove : Angle AOB And Angle APB are supplementary ( 180°)

                               or

                    ∠AOB + ∠APB = 180°

PROOF : Now in Δ OAP And Δ OBP, We have : ( both angles  

                                                                          having 90° angles )                                                              

PA = PB ( Tangents drawn from an external points are equal )

OA = OB ( Each equal to radius )

Op = OP ( Common )

∴ Δ OAP ≅ Δ OBP ( By SSS criterion of congruence )

⇒ ∠OPA = ∠OPB and ∠AOP = ∠BOP

⇒ ∠ APB = 2∠OPA and ∠AOB = 2 ∠AOP ----------- ( 1 )

But , ∠AOP = 90°-∠OPA  [ Triangle OAP is a 90° triangle ]

∴ 2∠AOP = 180°-2∠OPA

⇒ ∠AOB = 180°-∠APB     -------- [ USING Eq ( 1 ) ]

⇒  ∠AOB + ∠APB = 180°

HENCE PROVED...!!!

Answer 2

$\textbf{Answer is in Attachment !}$