Question

please answer quickly... I'm in very urgent for writing my assignment​

Answer 1

Answer:

In ΔPAC and ΔPBC

PA=PB

[length of tangents drawn from external point are equal ]

∠APC=∠BPC

[PA and PB are equally inclined to OP]

and PC=PC [Common]

So, by SAS criteria of similarity

ΔPAC≅ΔPBC

⇒AC=BC and ∠ACP=∠BCP

But ∠ACP+∠BCP=180

∘

∴∠ACP+∠BCP=90

∘

Hence, OP⊥AB