✔ 最佳答案
25a) ∠CDE = ∠ABC (Ext. ∠, cyclic quad.)
∠CDE = ∠EFG (Ext. ∠, cyclic quad.)
Hence ∠ABC = ∠EFG
b) ∠APC + ∠ABC = 180 (Opp. ∠s, cyclic quad.)
∠EPG + ∠EFG = 180 (Opp. ∠s, cyclic quad.)
∠APC + ∠ABC = ∠EPG + ∠EFG
∠APC = ∠EPG
∠APC + ∠CPE = ∠EPG + ∠CPE
∠APE = ∠CPG
26a) ∠FCE = ∠BCE = 34 (Equal chord, equal ∠ at circum)
b) ∠ABE = ∠ECF = 34 (∠ in alt. segment)
With ∠ABE = ∠DCE, ABCD is a cyclic quad. (Converse of ∠ in the same segment)
c) ∠ADE = ∠BCE = 34 (∠ in the same segment)
Hence ∠ABD = ∠ADB and so △ABCD is an isos. △
27a) Join OP, then △OMP and △ONP are congurent (RHS). Hence MP = NP.
b) AB = CD (Equal distance from centre, equal chord)
Also OM and ON bisects AB and BD resp. since they are perp. to each other.
Hence AM = MB = NC = ND
MB = NC
MB - MP = NC - NP
BP = CP