F3 四邊形的特性

26)aDC = AB (長方形性質) CR = AP (已知) DC - CR = AB - AP∴ DR = PB .......(1)AD = BC　(長方形性質) AS = CQ (已知) AD - AS = BC - CQ∴ SD = QB ........(2)∠ D = ∠ B = 90・ ......(3) (長方形性質) △ DSR 全等 △ BQP (SAS) 由(1)(2)(3)
bAP = CR (已知) AS = CQ (已知) ∠ A = ∠ C = 90・ (長方形性質) △ SAP 全等 △ QCR (SAS) SR = PQ (全等△,對應邊相等) SP = RQ (全等△,對應邊相等)∴ PQRS 是平衡四邊形 (兩組對邊相等)
28)P是AB中點 (已知) S是AC中點 (已知) PS = 1/2 BC (中點定理)PS ∕∕ BC (中點定理) Q是OB中點 (已知) R是OC中點 (已知) QR = 1/2 BC (中點定理)QR ∕∕ BC (中點定理)∴ PS = QR, PS ∕∕ QR∴ PQRS 是平衡四邊形 (一組對邊平衡且相等)
30)aH是BD中點 (平衡四邊形 性質) AB ∕∕ FH ....(1) (已知) DF = FA ......(2) (截線定理) 由(1)(2) FH = 1/2 AB .......(3) (中點定理)GH ∕∕ ED ......(4) (已知) 由(1)(4) AG = GE......(5) (截線定理) 由(1)(5) GH = 1/2 ED (中點定理) 考慮 △ GHF 和 △ GAB∠ FGH = ∠ BGA (對頂角) ∠ FHG = ∠ BAG (內錯角, AB ∕∕ FH)∠ HFG = ∠ ABG (內錯角, AB ∕∕ FH)∴ △ GHF 相似 △ GAB (AAA)∴ GH/GA = HF/AB (相似△,對應邊成比例) 由(3) FH = 1/2 AB∴ GH = 1/2 GA 由(5) GH = 1/2 ED
∴ ED = GA bAH = HC (平衡四邊形 性質) AG : GH = 2:1 (已證)AG : GH : HC = AG : GH : AG + GH = 2 : 1: 2+1AG : GH : HC = 2 : 1 : 3




2011-03-15 15:41:57 補充：
http://hk.knowledge.yahoo.com/question/question?qid=7009122601300