Two tangents PA and PB are drawn to a circle with center O from an external point P. Prove that ∠APB=2∠OAB [3 MARKS]
Concept: 1 Mark
Application: 2 Mark
OA⊥PA (tangent and the radius at the point of contact)
∠OAP=90∘
Let ∠OAB=x
∠PAB=90∘−x…(1)
PA = PB (tangent from same external point)
△APB is isosceles
∠PAB=∠PBA=90∘−x
In △APB, By angle sum property
∠APB + ∠PAB + ∠PBA=180∘
∠P+90∘−x+90∘−x=180∘
∠P+180∘−2x=180∘
∠P=2x
∠APB=2x
i.e., ∠APB=2∠OAB