較難の演繹幾何

2.
∠ABD = ∠CDB (內錯角 , AB//DC)
∠ADB = ∠CBD (內錯角, AD//BC)
BD = BD (公共邊)
ΔABD ≡ ΔCDB (ASA)


3.
BC = FE ( 已知
BC + CF = FE + CD (公理)
所以BF = EC

BF = EC (已證)
∠ABF = ∠DEC (內錯角, BA//DE)
∠AFB = ∠DCE (內錯角, AF//CD)
ΔABF ≡ ΔDEC (ASA)


4.(a)
∠ABC + ∠ACB + 90° = 180o (ΔABC 內角和)
∠DCE + ∠ACB + 90° = 180o (直線同側鄰角)
所以 ∠ABC = ∠DCE (公理)

∠ABC = ∠DCE (已證)
AB = DC (已知)
∠BAC = ∠CDE = 90° (已知)
ΔABC ≡ ΔDCE (ASA)

4.(b)
ΔABC ≡ ΔDCE (已證)
BC = CE (全等Δ對應邊)
所以 ΔBCE 是等腰三角形 (兩邊 BC、CE 等長)


5.(a)
AB = AC ( 已知)
BM = CM (已知)
AM = AM (公共邊)
ΔBAM ≡ ΔCAM (SSS)
∠BAM = ∠CAM (全等Δ對應角)

5.(b)
∠BAM = ∠CAM (已證)
AB = AC (已知)
AN = AN (公共邊)
ΔBNA ≡ ΔCNA (SAS)
BN = CN (全等Δ對應邊)


6.(a)
∠BAC = ∠BDC ( 已知)
∠CAD = ∠ADB (已知)
∠BAC + ∠CAD = ∠BDC + ∠ADB (公理)
所以∠BAD = ∠CDA

∠BAD = ∠CDA (已證)
∠ADB = ∠CAD (已知)
AD = AD (公共邊)
ΔABD ≡ ΔDCA (ASA)

6(b)
ΔABD ≡ ΔDCA (已證)
AB = DC (全等Δ對應邊)
AC = DB (全等Δ對應邊)
BC = CB (公共邊)
ΔABC ≡ ΔDCB (SSS)


7.
∠ADE = ∠ABC ( 同位角, DE//BC)
∠AED = ∠ACB (同位角, DE//BC)
∠A = ∠A (公共角)
ΔADE ~ ΔABC (AAA)

AD/AB = AE/AC = DE/BC (相似Δ對應邊)
故 ΔABC 中任兩邊相等，ΔADE 的兩對應邊亦相等。(相似Δ對形邊)


8.(a)
∠ABD = ∠CED ( 內錯角, AB//EC)
∠BAD = ∠ECD (內錯角, AB//EC)
AB = EC (已知)
ΔABD ≡ ΔCED (ASA)
BD = DE (全等Δ對應邊)

8.(b)
ΔABD ≡ ΔCED (已證)
BD = ED (全等Δ對應邊)

ΔABD ≡ ΔCED (已證)
∠ABD = ∠CED (全等Δ對應角)
∠ABD = ∠CBD (已知)
所以 ∠CBD = ∠CED (公理)
CB = CE (等腰ΔCBE兩底角)
BD = ED (已證)
ΔBDC ≡ ΔEDC (SAS)

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