Bài 3:
a)
$\Delta ADE$ có $AD\,\,||\,\,BK$
$\to \dfrac{AD}{BK}=\dfrac{AE}{EK}$ ( hệ quả định lý Ta – let )
$\to AD=\dfrac{AE\,.\,BK}{EK}=\dfrac{12.8}{6}=16\,\,\left( cm \right)$
b)
$\Delta ABE$ có $AB\,\,||\,\,DG$
$\to \dfrac{AE}{EG}=\dfrac{EB}{ED}$ ( định lý Ta – let )
$\Delta ADE$ có $AD\,\,||\,\,BK$
$\to \dfrac{EK}{AE}=\dfrac{EB}{ED}$ ( định lý Ta – let )
$\to \dfrac{AE}{EG}=\dfrac{EK}{AE}$
$\to A{{E}^{2}}=EK.EG$
c)
$\Delta ADE$ có $AD\,\,||\,\,BK$
$\to \dfrac{AE}{AK}=\dfrac{DE}{BD}$ ( định lý Ta – let )
$\Delta ABE$ có $AB\,\,||\,\,DG$
$\to \dfrac{AE}{AG}=\dfrac{BE}{BD}$ ( định lý Ta – let )
$\to \dfrac{AE}{AK}+\dfrac{AE}{AG}=\dfrac{DE}{BD}+\dfrac{BE}{BD}=\dfrac{BD}{BD}=1$
Bài 4:
$\,\,\,\,\,\,\,\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}$
$\Leftrightarrow \dfrac{x-2014+2011}{2011}+\dfrac{x-2014+2012}{2012}=\dfrac{x-2014+2}{2}+\dfrac{x-2014+3}{3}$
$\Leftrightarrow \dfrac{x-2014}{2011}+1+\dfrac{x-2014}{2012}+1=\dfrac{x-2014}{2}+1+\dfrac{x-2014}{3}+1$
$\Leftrightarrow \dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0$
$\Leftrightarrow \left( x-2014 \right)\left( \dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3} \right)=0$
$\Leftrightarrow x-2014=0$ vì $\left( \dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\,\,\ne \,\,0 \right)$
$\Leftrightarrow x=2014$
