圓形証明~圓形証明~

連 PQ , ∠RPQ = ∠RDQ (∠s in the same segment 同弧所對圓周角相等) ...... (1)∠ASQ = ∠RDQ + ∠DQS (ext. ∠ of Δ △外角) ...... (2)
又∠ASQ
= ∠APQ (∠s in the same segment 同弧所對圓周角相等)
= ∠APR + ∠RPQ ...... (3)
以 (1) 代入 (2) :
∠ASQ = ∠RPQ + ∠DQS ...... (4)
以 (3) 代入 (4) :
∠APR + ∠RPQ = ∠RPQ + ∠DQS
∠APR = ∠DQS
∠APR = 180° - ∠SQC (adj. ∠s on st. line 直線上的鄰角)
70° = 180° - ∠SQC
∠SQC = 110°

Join A to C, P to Q, B to D.

∠APQ = ∠BDQ (ext ∠ of cyclic quad)
∠APQ +∠ACQ = 180° (opp ∠s of cyclic quad)
∴ ∠BDQ+∠ACQ = 180°
∴ AC // BD (int ∠s supp )

∠BDR = ∠APR = 70° (ext ∠ of cyclic quad)
∠RAC = ∠BDR = 70° (alt ∠s, AC // BD)
∠SQC = 180° - ∠RAC (opp ∠s of cyclic quad)
= 110°

請問你要證明什麼呢？