Square ABCD is inscribed in circle O, P is a point on inferior arc BC, AP intersects BD with Q, QP = Qo, and QD / Qo is obtained= We haven't learned similarity and trigonometric functions yet

Square ABCD is inscribed in circle O, P is a point on inferior arc BC, AP intersects BD with Q, QP = Qo, and QD / Qo is obtained= We haven't learned similarity and trigonometric functions yet

Connecting BP, DP, Op, from equal arc and equal angle, it can be concluded that: ∠ APD = ∠ abd = 45 °
∵OP=OD
∴∠ODP=∠OPD
∵QP=QO
∴∠QPO=∠POQ=∠ODP+∠OPD=2∠ODP
∴∠APD=3∠ODP=45°
∴∠ODP=15°
∴BP=BDsin15°=√2 AD(√6-√2)/4=AD(√3-1)/2
△AQD∽△BQP
QD/QP=AD/BP=2/(√3-1)=√3+1
∴QD/QO=QD/QP=√3+1