toán 8

N

nhungle201

P

pinkylun

a) $\triangle{AFC}$~$\triangle{AEB}$ vì:

$\hat{A}$ chung

$\widehat{AFC}=\widehat{AEB}=90^o$

$=>\dfrac{AF}{AE}=\dfrac{AC}{AB}$

$=>AF.AB=AC.AE$

b) $\triangle{AEF}$~$\triangle{ABC}$ (c-g-c)

Vì :$\hat{A}$ chung

$\dfrac{AE}{AF}=\dfrac{AB}{AC}$ (cmt)

c) Vẽ $HI \perp BC(I\ \in \ BC )$

Dể cm $\triangle{BHI}$~$\triangle{BCE}$

$=>\dfrac{BH}{BC}=\dfrac{BI}{BE}$

$=>BH.BE=BC.BI$ :D

Tương tự:

$\triangle{CIH}$~$\triangle{CFB}$

$=>\dfrac{CI}{CF}=\dfrac{CH}{CB}$

$=>CH.CF=BC.CI$ :)

Cộng hai vế :D và :) $=>đpcm$ :D
 
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