Toán nâng cao đại số 8

V

viethoang345

H

hien_vuthithanh

a/ chứng minh [TEX]\triangle \ [/TEX] CDF= [TEX]\triangle \ [/TEX] BCE(2cgv)
\Rightarrow $\widehat{CFD}= \widehat{BEC}$
\Rightarrow $\widehat{CFD}+\widehat{MCF}$=$\widehat{MCF}+ \widehat{BEC}$
\Rightarrow$ \widehat{CMF}=90^0$

b/ $\dfrac{CM.CE}{CF}$=a \Leftrightarrow CM.DF=CF.CD(=2$S_{DCF}$) (Vì EC=DF)
\Rightarrow dpcm
 
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D

dien0709

G,H trung điểm DC,DA, nối AG,BH.Dùng đl đường TB dễ thấy:

[TEX] CM=2/5CE=\frac{a}{\sqrt[]{5}}; DM=4/5DF=\frac{2a}{\sqrt[]{5}}[/TEX]

[TEX]=>S_{DMC}=\frac{a^2}{5}[/TEX]
 
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