✔ 最佳答案
i)
ÐMNP = ÐPNK (common Ðs)
ÐMPN = ÐPKN (Ð in alt. segment)
ÐPMN = ÐKPN (The 3rd Ðs of 2 Δs with 2 Ðs equal)
ΔMPN ~ ΔPKN (AAA)
NP/NK = NM/NP (corr. sides of similar Δs)
Hence, NP² = (NK)(NM) (rearrangement)
ii)
ON = NM (2 tangents from a pt. to a circle)
RM // ON (given)
ÐKRM = ÐKON (corr. Ðs, //)
ÐKMR = ÐKNO (corr. Ðs, //)
ÐRKM = ÐOKN (common Ðs)
ΔKRM ~ KON (AAA)
RM/ON = KM/KN (corr. sides of similar Δs)
RM/NM = KM/KN (substitution)
NP = NM (2 tangents from a pt. to a circle)
MS // NP (given)
ÐKMS = ÐKNP (corr. Ðs, //)
ÐKSM = ÐKPN (corr. Ðs, //)
ÐMKS = ÐNKP (common Ðs)
ΔKMS ~ KNP (AAA)
MS/NP = KM/KN (corr. sides of similar Δs)
MS/NM = KM/KN (substitution)
Compare RM/NM = KM/KN and MS/NM = KM/KN
RM/NM = MS/NM (axiom)
Hence, RM = MS (axiom)
2011-09-09 13:30:32 補充：
I am sorry to make a mistake in interpretation of the graph. The amendment of Part ii) is as follows.
2011-09-09 13:31:01 補充：
..... (cont'd)
Join OM.
∠MNO = ∠ONK (common ∠s)
∠MON = ∠OKN (∠ in alt. segment)
∠OMN = ∠KON (The 3rd ∠s of 2 Δs with 2 ∠s equal)
ΔMON ~ ΔOKN (AAA)
NO/NK = NM/NO (corr. sides of similar Δs)
Hence, NO² = (NK)(NM) (rearrangement)
2011-09-09 13:31:43 補充：
..... (cont'd)
In Part i), it has been proved that NP² = (NK)(NM).
Compare NP² = (NK)(NM) and NO² = (NK)(NM)
NP² = NO² (axiom)
Hence NP = NO
2011-09-09 13:32:15 補充：
..... (cont'd)
RM // ON (given)
∠KRM = ∠KON (corr. ∠s, //)
∠KMR = ∠KNO (corr. ∠s, //)
∠RKM = ∠OKN (common ∠s)
ΔKRM ~ KON (AAA)
RM/ON = KM/KN (corr. sides of similar Δs)
But NP = NO (proven)
Hence, RM/NP = KM/KN
2011-09-09 13:32:47 補充：
..... (cont'd)
MS // NP (given)
∠KMS = ∠KNP (corr. ∠s, //)
∠KSM = ∠KPN (corr. ∠s, //)
∠MKS = ∠NKP (common ∠s)
ΔKMS ~ KNP (AAA)
MS/NP = KM/KN (corr. sides of similar Δs)
Compare RM/NP = KM/KN and MS/NP = KM/KN
RM/NP = MS/NP (axiom)
Hence, RM = MS (axiom)
