Đáp án:bài 1 x=2,1cm
$Y=\frac{20}{3}cm$
CE=8cm
DE=4,5cm
BE=9cm
Giải thích các bước giải:
$ \Delta JGH~\Delta JKI(G.G)$
=>$ \frac{JG}{JK}=\frac{JH}{JI}=\frac{GH}{KI}$
=>$ x= JH=\frac{JG.JI}{JK}= \frac{3×3.5}{5}= 2,1$
$ y=IK=\frac{GH.JK}{JG}=\frac{ 4×5}{3}=\frac{20}{3}$
$\Delta ADE~\Delta ABC$
$\widehat{A} chung$
$\widehat{ADE}=\widehat{ABC}( Đồng vị)$
=>$ \frac{ AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$
=>$ AC=\frac{AE.AB}{AD}=\frac{ 6×7}{3}=14$
=> EC=AC-AE=14-6=8
$DE= \frac{AE.BC}{AC}=\frac{3.21}{14}= 4,5$
$\Delta CFE~\Delta CBA$
=> $\frac{CF}{CB}=\frac{CE}{CA}$
=> $CF=\frac{CB.CE}{CA}=\frac{ 21×8}{14}$
= 12
=> BE=BC-BF=21-12=9
