sossssssssssssssssssssssssss

Giải thích các bước giải:

 a.Xét $\Delta CGH,\Delta CEB$ có:

$\widehat{CGH}=\widehat{CEB}(=90^o)$

Chung $\hat C$

$\to \Delta CGH\sim\Delta CEB(g.g)$

b.Xét $\Delta BGH,\Delta BDC$ có:

Chung  $\hat B$

$\widehat{BGH}=\widehat{BDC}(=90^o)$

$\to \Delta BGH\sim\Delta BDC(g.g)$

c.Từ câu a $\to \dfrac{CH}{CB}=\dfrac{CG}{CE}\to CH\cdot CE=CG\cdot CB$

                 b $\to \dfrac{BH}{BC}=\dfrac{BG}{BD}\to BH\cdot BD=BG\cdot BC$

$\to BH\cdot BD+CH\cdot CD=BG\cdot BC+CG\cdot BC=BC^2$

$\to BH\cdot BD+CH\cdot CD-BC^2=0$