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標題:
f3 mth questions!!!急~星期二要交!
發問:
5. In the figure,AB = 12cm ,BC = 10cm ,CD = 15cm ,DA = 27cm and AC = 18cm .Prove that AC bisects
最佳解答:
5. Given: AB = 12 cm, BC = 10 cm, CD = 15 cm, DA = 27 cm and AC = 18 cm To prove: AC bisects ∠BAD Proof: AB/AC = 12/18 = 2/3 (calculation) AC/AD = 18/27 = 2/3 (calculation) BC/CD = 10/15 = 2/3 (calculation) AB/AC = AC/AD = BC/CD (axiom) ΔABC ~ ΔACD (3 sides in proportion) ∠BAC = ∠CAD (corr. ∠s of similar Δs) Hence, AC bisects ∠BAD. 13. Given: BE⊥AC, ACD is a st. line, and ∠CBD = ∠BAC To prove: (a) ΔABC ~ ΔBDC (b) BE2 = AC x DC - EC2 (a) ∠CAB = ∠CBD (given) ∠C = ∠C (common ∠) Hence, ∠ABC = ∠BDC (the 3rd ∠ of Δs with 2∠s equal) ΔABC ~ ΔBDC (AAA) (b) BE⊥EC (given) In ΔBEC: BC2 = BE2 + EC2 ...... (*) From (a), ΔABC ~ ΔBDC (proved) BC/DC = AC/BC (corr. sides, similar Δs) BC2 = AC x DC (rearrangement) ...... (#) Compare (*) and (#): BC2 = BC2 BE2 + EC2 = AC x DC (axiom) BE2 = AC x DC - EC2 (rearrangement)
其他解答:
f3 mth questions!!!急~星期二要交!
發問:
5. In the figure,AB = 12cm ,BC = 10cm ,CD = 15cm ,DA = 27cm and AC = 18cm .Prove that AC bisects
最佳解答:
5. Given: AB = 12 cm, BC = 10 cm, CD = 15 cm, DA = 27 cm and AC = 18 cm To prove: AC bisects ∠BAD Proof: AB/AC = 12/18 = 2/3 (calculation) AC/AD = 18/27 = 2/3 (calculation) BC/CD = 10/15 = 2/3 (calculation) AB/AC = AC/AD = BC/CD (axiom) ΔABC ~ ΔACD (3 sides in proportion) ∠BAC = ∠CAD (corr. ∠s of similar Δs) Hence, AC bisects ∠BAD. 13. Given: BE⊥AC, ACD is a st. line, and ∠CBD = ∠BAC To prove: (a) ΔABC ~ ΔBDC (b) BE2 = AC x DC - EC2 (a) ∠CAB = ∠CBD (given) ∠C = ∠C (common ∠) Hence, ∠ABC = ∠BDC (the 3rd ∠ of Δs with 2∠s equal) ΔABC ~ ΔBDC (AAA) (b) BE⊥EC (given) In ΔBEC: BC2 = BE2 + EC2 ...... (*) From (a), ΔABC ~ ΔBDC (proved) BC/DC = AC/BC (corr. sides, similar Δs) BC2 = AC x DC (rearrangement) ...... (#) Compare (*) and (#): BC2 = BC2 BE2 + EC2 = AC x DC (axiom) BE2 = AC x DC - EC2 (rearrangement)
其他解答:
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16: (a) △ABC ~ △BDC C=C(common <)
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