(F.4-f.5)circle angle問題

14a) ∠ABC = ∠ACP = x (∠in alt. segment)
∠ABD = ∠ADP = y (∠in alt. segment)
∴ ∠CBD = ∠ABC + ∠ABD = x + y
b)∠CPD = 180° - ∠PCD - ∠PDC = 180° - x - y (∠ sum of Δ)
∵ ∠CPD + ∠CBD = (180° - x - y) + (x + y) = 180°
∴ B,C,P,D are concyclic (opp. ∠s supp.) Alternatively :∠BCP + ∠BDP = 90° + 90° (tangent⊥radius) = 180°
∴ B,C,P,D are concyclic (opp. ∠s supp.)
c) ∠ABC = x (Proved in part a) )By b) , B,C,P,D are concyclic ,
Hence ∠ABC = ∠ADP (∠s in the same segment)
i.e. x = y.
16ai)∠DQA
=∠CAQ + b (ext. ∠ of Δ )
= a + b (equal arcs, equal ∠s)
aii)∠ABP
= a + ∠ABD
= a + b (∠s in the same segment)
= ∠DQA (Proved in ai)
∴ ABRQ is a cyclic quadrilnteral. (converse of ext.∠= int. opp.∠)
b)a = ∠DBC /2 (equal arcs, equal ∠s) = 70/2 = 35°∠BDC
= ∠CAB (∠s in the same segment)
= ∠QAB - ∠QAC
= ∠QAB - a (equal arcs, equal ∠s)
= ∠PRQ - a (ext.∠= int. opp.∠)
= 80° - 35°
= 45°