Giải thích các bước giải:
Bài 9:
Xét $\Delta ADC,\Delta BEC$ có:
Chung $\hat C$
$\widehat{ADC}=\widehat{BEC}(=90^o)$
$\to \Delta ADC\sim\Delta BEC(g.g)$
Xét $\Delta AHE,\Delta BHD$ có:
$\widehat{AHE}=\widehat{BHD}$
$\widehat{AEH}=\widehat{HDB}(=90^o)$
$\to\Delta AHE\sim\Delta BHD(g.g)$
$\to \dfrac{HA}{HB}=\dfrac{HE}{HD}$
$\to HA.HD=HB.HE$
Bài 10:
Xét $\Delta ABE,\Delta ACF$ có:
Chung $\hat A$
$\widehat{AEB}=\widehat{AFC}(=90^o)$
$\to\Delta ABE\sim\Delta ACF(g.g)$
$\to \dfrac{AB}{AC}=\dfrac{AE}{AF}$
$\to \dfrac{AB}{AE}=\dfrac{AC}{AF}$
Mà $\widehat{BAC}=\widehat{EAF}$
$\to\Delta ABC\sim\Delta AEF(c.g.c)$
$\to \dfrac{BC}{EF}=\dfrac{AB}{AE}$
$\to AE.BC=AC.EF$