a,
$\Delta$ ABC vuông tại A có BC= $\sqrt{AB^2+ AC^2}$= 10 cm
BD phân giác => $\frac{AD}{DC}= \frac{AB}{BC}= \frac{6}{10}= \frac{4}{5}$
b,
$\Delta$ ADB ~ $\Delta$ ECB (g.g) vì $\widehat{CEB}= \widehat{DAB}= 90^o$, $\widehat{EBC}= \widehat{ABD}$
=> $\frac{BD}{AD}= \frac{BC}{EC}$
=> BD.EC= AD.BC
c,
$\widehat{CEH}= 180^o- \widehat{CEB} - \widehat{HEB}= 90^o - \widehat{HEB}$ (*)
$\widehat{HBE}= 180^o - \widehat{EHB} -\widehat{HEB}= 90^o - \widehat{HEB}$ (**)
(*)(**) => $\widehat{CEH}= \widehat{HBE}$
$\Delta$ CEH ~ $\Delta$ CBE (g.g) vì có chung $\widehat{ECH}$, $\widehat{CEH}= \widehat{HBE}$
=> $\frac{EC}{ED}= \frac{EB}{EC}$
=> $EC^2= ED.EB$ (1)
$\frac{EC}{CH}= \widehat{CB}{EC}$
=> $EC^2= CH.CB$ (2)
(1)(2) => $ED.EB= CH.CB$
