5. In the figure,AB = 12cm ,BC = 10cm ,CD = 15cm ,DA = 27cm and AC = 18cm .Prove that AC bisects <BAD.
16.In the figure,BE丄AC. D is a point on AC such that <CBD = <BAC. Prove that
(a) △ABC ~ △BDC,
(b) BE^2 = AC x DC - EC^2.
因為係有圖啲,但我up唔到係lee到,所以要留低msn or e-mail
我會sd圖比你地,到時你再係lee到答番,就即有15分!!!
求step,thx~~
一定要有完整啲step~~(一定要有(< sum of △)lee d公式!!)唔好亂做,一定要岩同埋要清楚(要睇得明你寫咩)!!!!thx~~
f3 mth questions!!!急~星期二要交!
2011-01-24 9:43 am
回答 (2)
2011-01-24 11:53 am
✔ 最佳答案
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)
參考: micatkie
2011-01-25 2:17 am
16:
(a) △ABC ~ △BDC
C=C(common <)
(a) △ABC ~ △BDC
C=C(common <)
收錄日期: 2021-04-13 17:47:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110124000051KK00102