Two chords AC and BD of a circle intersect each other at the point O. If two tangents drawn at the points A and B intersect each other at the point P and two tangents drawn at the points C and D intersect at the point Q, let us prove that ∠P + ∠Q = 2 ∠BOC.

Formula used.

⇒ Isosceles triangle property

If 2 angles of triangle are equal then their corresponding sides are also equal ⇒ Perpendicular drawn through tangent pass through centre

Solution

Mark centre of circle to be X

Join AX, CX, BX, DX radius of circle

In Δ BDX

As XB = XD

By isosceles triangle property

∠DBX = ∠BDX

In Δ ACX

As XA = XC

By isosceles triangle property

∠ACX = ∠CAX

In quadrilateral APBO

∠OAP = ∠XAP - ∠XAC

= 90° - ∠XAC ∵ (AP is tangent at point A)

∠OBP = ∠XBP + ∠XBD

= 90° + ∠XBD ∵ (BP is tangent at point B)

By angle sum property

∠OBP + ∠P + ∠OAP + ∠AOB = 360°

∠AOB = 360° - ∠P - ∠OBP - ∠OAP

= 360° - ∠P – [90° - ∠XAC ] – [90° + ∠XBD]

= 180° - ∠P + ∠XAC – ∠XBD

In quadrilateral DQCO

∠ODQ = ∠XDQ - ∠XDB

= 90° - ∠XDB ∵ (AP is tangent at point A)

∠OCQ = ∠XCQ + ∠XCA

= 90° + ∠XCA ∵ (BP is tangent at point B)

By angle sum property

∠ODQ + ∠Q + ∠OCQ + ∠DOC = 360°

∠DOC = 360° - ∠Q - ∠ODQ - ∠OCQ

= 360° - ∠Q – [90° - ∠XDB ] – [90° + ∠XCA]

= 180° - ∠Q + ∠XDB – ∠XCA

As ∠DBX = ∠BDX and ∠ACX = ∠CAX

= 180° - ∠Q + ∠XBD – ∠XAC

∠AOB + ∠DOC

= 180° - ∠P + ∠XAC – ∠XBD + [180° - ∠Q + ∠XBD – ∠XAC]

= 360° - ∠P - ∠Q

∠BOC = ∠AOD ∵ (Vertically opposite angle)

[∠BOC + ∠AOD] + [∠AOB + ∠DOC] = 360°

2∠BOC + [360° - ∠P - ∠Q] = 360°

2∠BOC = ∠P + ∠Q