✔ 最佳答案
1.
因N, C, B, P 共圓, CP 為直徑,
∴∠CNP = 90°, ∠CND + ∠PNA = 90°
又∠CND + ∠NCD = 90° 故 ∠NCD = ∠PNA
CN = PN, 得ΔCDN ≡ΔANP (AAS)
令 AP = X, BP = Y, 則DN = X, AN = CD = X + Y, AD = 2X + Y
四邊形CNPB面積 = 長方形ABCD面積 - ΔCDN面積 -ΔANP面積
= (X+Y)(2X+Y) - X(X+Y)/2 - X(X+Y)/2 = (X+Y)^2 = 7
∴AB = CD = X+Y = √7
2.
連CF交AB於M, 連CG交 DE於N, 則M、N分別為ΔABC、ΔCDE之外心,
故 CM= AM= 6, CN = DN = 6,
得 CF = CG = 6X2/3 = 4
因∠1 = ∠2, ∠ACB = ∠ECD = 90° , AB = DE, ∴ΔACB≡ΔDCE (AAS)
得BC = EC, 故 ΔBCM≡ΔECN, 且 ΔACM≡ΔDCN (SSS)
∠BCM = ∠ECN 且∠ACM = ∠DCN
故∠ACM + ∠ECN = 90°, 即ΔFCG為等腰直角Δ
∴FG = 4√2