Question

In a circle with center o, chord AB = Chord
PQ show that angle AOB = angle POQ​

Answer 1

Answer:

In △POQ

Let ∠OPQ=x

WKT OP=OQ=radius

⟹∠OPQ+∠OQP+POQ=180

∠POQ=180–2x

Also, radius is perpendicular to tangent

∠OPR=90

∠QPR=∠OPR−∠OPQ

∠RPQ=90–x=

2

∠POQ

2∠RPQ=∠POQ