Two tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that PR = PA + AQ.
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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
Step-by-step explanation:
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