Giải thích các bước giải:
a.Ta có:
$BC^2=AB^2+AC^2\to \Delta ABC$ vuông tại $A$
b.Ta có: $\widehat{AEK}=\widehat{HEC},\widehat{EAK}=\widehat{EHC}=90^o$
$\to \Delta AEK\sim\Delta HEC(g.g)$
$\to\dfrac{AE}{HE}=\dfrac{EK}{EC}$
$\to EA.EC=EH.EK$
c.Ta có $AE=AC-CE=5$
Ta có: $\widehat{CHE}=\widehat{CAB}=90^o,\widehat{ECH}=\widehat{ACB}$
$\to \Delta CHE\sim\Delta CAB(g.g)$
$\to\dfrac{EH}{AB}=\dfrac{CE}{CB}$
$\to EH=\dfrac{CE\cdot AB}{CB}=9$
Lại có:
$EA\cdot EC=EH\cdot EK$
$\to EK=\dfrac{EA\cdot EC}{EH}$
$\to EK=\dfrac{25}{3}$
$\to KH=EK+EH=\dfrac{52}{3}$
$\to \dfrac{S_{BCE}}{S_{BCK}}=\dfrac{\dfrac12\cdot BC\cdot EH}{\dfrac12\cdot BC\cdot HK}=\dfrac{EH}{KH}=\dfrac{27}{52}$