Giải thích các bước giải:
a.Ta có $\Delta ABC$ đều $\to AB=AC$
$\to \widehat{ADB}=\widehat{ADC}$
b.Ta có:
$\widehat{BAE}=\widehat{BAD}$
$\widehat{ABE}=\widehat{ABC}=\widehat{ADC}=\widehat{ADB}$
$\to\Delta ABE\sim\Delta ADB(g.g)$
$\to \dfrac{AB}{AD}=\dfrac{AE}{AB}$
$\to AB^2=AD.AE$
c.Trên $DA$ lấy điểm $F$ sao cho $DF=DC$
$\to \Delta DCF$ cân tại $D$
Mà $\widehat{FDC}=\widehat{ADC}=\widehat{ABC}=60^o$
$\to\Delta DCF$ đều
Xét $\Delta ACF,\Delta BDC$ có:
$AC=BC$
$\widehat{ACF}=\widehat{ACB}-\widehat{BCF}=60^o-\widehat{BCF}=\widehat{DCF}-\widehat{BCF}=\widehat{BCD}$
$CD=CF$
$\to \Delta ACF=\Delta BCD(c.g.c)$
$\to BD=AF$
$\to AD=AF+FD=BD+CD$
d.Ta có:
$\dfrac{1}{BD}+\dfrac{1}{CD}=\dfrac{BD+CD}{BD.CD}=\dfrac{AD}{DB.DC}$
Xét $\Delta BDE, \Delta ADC$ có:
$\widehat{DBE}=\widehat{DAC},\widehat{BDE}=\widehat{BDA}=\widehat{ADC}$
$\to \Delta BDE\sim\Delta ADC(g.g)$
$\to \dfrac{BD}{AD}=\dfrac{DE}{DC}$
$\to BD.DC=DA.DE$
$\to \dfrac{AD}{DB.DC}=\dfrac1{DE}$
$\to \dfrac{1}{BD}+\dfrac1{CD}=\dfrac1{DE}$
