Question

圆形证明~圆形证明~

Answer 1

更新1:

本图不照比例绘画

连 PQ
∠RPQ = ∠RDQ (∠s in the same segment 同弧所对圆周角相等) ...... (1)∠ASQ = ∠RDQ + ∠DQS (ext. ∠ of Δ △外角) ...... (2) 又∠ASQ = ∠APQ (∠s in the same segment 同弧所对圆周角相等) = ∠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 直线上的邻角) 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°
请问你要证明什么呢？