Question

From a point P outside a circle, with centre O. tangents PA and PB are drawn as following fig., Prove that ∠ AOP = ∠ BOP and OP is the perpendicular bisector of AB. 

Answer 1

In Δ AOP =Δ BOP 

AP = PB (lengths of tangents drawn from and external point to a circle are equal) 

OP = PO (common) 

∠ PAO = ∠ PBO = 90 ° (radius is ⊥  to tangent at the point of contact) 

∴  Δ AOP ≅  Δ BOP   {By RHS} 

Δ AOP =  Δ BOP {By CPCT} 

In Δ AMO and Δ BMO 

AO = OB (radius of same circle) 

∠ MOA = ∠ MOB {Proved above) 

OM = MO {Common) 

∴ Δ AMO ≅  Δ BMO {By CPCT} 

∠ AMO = ∠ BMO 

∠ AMO + ∠ BMO = 180° 

∴ 2 ∠ AMO = 180° 

∠ BMO = ∠ AMO =  90° 

Hence, OP is the perpendicular bisector of AB.