急!急! F4 Properties Circles q7

25)∠PRQ = ∠POQ/2 = 62/2 = 31° (∠at centre twice ∠ at ☉ce)∠OPQ = (180 - 62)/2 = 59° = ∠RPS
Let PO meet ☉ce at p , Pp = Pp (common)
Rp = Qp (equal ∠s, equal chords)
∴ △PpR ≌ △PpQ (S.A.S.) , So PR = PQ
∠OPQ = ∠RPS (proved)
PS = PS (common)
∴ △PRS ≌ △PQS (S.A.S.)So ∠RSP = ∠QSP , but ∠RSP + ∠QSP = 180°
Hence ∠RSP = ∠QSP = 90°So PQ² = 1.2² + 2²
PQ = √5.44 ≈ 2.33 cm

26)CO = CO (common)
BC = CD (given)
OD = OB (radii)
∴ △OCD ≌ △OCB (S.S.S.)So ∠DCO = ∠BCO
BC = CD (given)
Let CO meet BD at c , Cc = Cc (common)
∴ △DCc ≌ △BCc (S.A.S.)So ∠CcD = ∠CcB , but ∠CcD + ∠CcB = 180°
Hence ∠CcD =∠CcB = 90°So ∠OBC = ∠OCB = 180° - 90° - ∠CBD = 90° - ∠CBD
∠BCD = 180° - 2∠CBD

18° + ∠OBC + ∠BCD = 180° (∠sum of Δ )
18° + (90° - ∠CBD) + (180° - 2∠CBD) = 180°
108° = 3∠CBD
∠CBD = 36°

27a)2∠BRP = ∠BOP (∠at centre twice ∠ at ☉ce)
2∠BRP = ∠BOC/2 = ∠BAC(∠at centre twice ∠ at ☉ce)
2∠BRP = 180 - 38 - 72 = 70°
∠BRP = 35°b)∠RAC
= ∠RAB + ∠BAC
= ∠RAB + 70°
= (180° - ∠ARB)/2 + 70°
= (180° - (180°-72°))/2 + 70° (int.∠s supp.)
= 106°c)∠BRQ
= ∠BOQ / 2 (∠at centre twice ∠ at ☉ce)
= (∠BOC + ∠COQ) / 2
= (2 * 70° + ∠COA/2) / 2
= (140° + ∠ABC) / 2
= (140° + 38°) / 2
= 89°


2012-02-25 17:45:29 補充：
Q26 , the best way of find ∠OBC as follows :

∠OBC
= 180° - ∠ADC (int.∠s supp.)
= 180° - ∠ADB - ∠CDB
= 180° - 90° - ∠CBD
= 90° - ∠CBD

No need to prove ∠CcD =∠CcB = 90°.