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在圖中, AB,EF,DC都垂直於BC,證明
a) 三角形CEF ~ 三角形CAB
b) 三角形BEF ~ 三角形BDC
c) (1/a) + (1/b) = 1/c
a同b我識做,但
唔識做c題...請高手指點下...thx!
f.3 math ~ 15分
2008-08-30 11:42 pm
回答 (3)
2008-08-31 12:03 am
✔ 最佳答案
c:b = BF : (BF+CF) (corr. sides ~ triangles) - (A)c:a = CF : (BF+CF) (corr. sides ~ triangles) - (B)
(A) + (B)
c/b + c/a = BF / (BF+CF) + CF / (BF+CF)
c ( 1/b + 1/a ) = (BF + CF )/(BF + CF)
c ( 1/b + 1/a ) = 1
(1/a) + (1/b) = 1/c
2008-08-31 12:02 am
c)Let BF = m and FC = n.
In triangle ABC and triangle EFC, by similar triangle,
c/n = a/(m + n)
cm + cn = an
cm = an - cn = n(a - c)
m/n = (a-c)/c..........................(1)
For triangle DBC and triangle EBF, similarly,
m/c = (m+n)/b
bm = cm + cn
m(b - c) = cn
m/n = c/(b-c).......................(2)
(1) = (2)
(a-c)/c = c/(b -c)
(a-c)(b-c) = c^2
ab - cb - ac + c^2 = c^2
ab = cb + ac
All terms divided by abc, we get
1/c = 1/a + 1/b.
In triangle ABC and triangle EFC, by similar triangle,
c/n = a/(m + n)
cm + cn = an
cm = an - cn = n(a - c)
m/n = (a-c)/c..........................(1)
For triangle DBC and triangle EBF, similarly,
m/c = (m+n)/b
bm = cm + cn
m(b - c) = cn
m/n = c/(b-c).......................(2)
(1) = (2)
(a-c)/c = c/(b -c)
(a-c)(b-c) = c^2
ab - cb - ac + c^2 = c^2
ab = cb + ac
All terms divided by abc, we get
1/c = 1/a + 1/b.
2008-08-30 11:57 pm
BC/a = FC/c
BC/b = BF/c
BC/a + BC/b = FC/c + BF/c
BC( 1/a + 1/b) = (FC +BF)/c
BC( 1/a + 1/b) = BC/c
1/a + 1/b = 1/c
2008-08-30 15:58:42 補充：
BC/a = FC/c ....(1)
BC/b = BF/c ....(2)
(1) + (2), BC/a + BC/b = FC/c + BF/c
BC( 1/a + 1/b) = (FC +BF)/c
BC( 1/a + 1/b) = BC/c
1/a + 1/b = 1/c
BC/b = BF/c
BC/a + BC/b = FC/c + BF/c
BC( 1/a + 1/b) = (FC +BF)/c
BC( 1/a + 1/b) = BC/c
1/a + 1/b = 1/c
2008-08-30 15:58:42 補充：
BC/a = FC/c ....(1)
BC/b = BF/c ....(2)
(1) + (2), BC/a + BC/b = FC/c + BF/c
BC( 1/a + 1/b) = (FC +BF)/c
BC( 1/a + 1/b) = BC/c
1/a + 1/b = 1/c
參考: me, e個易睇d
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