Giải thích các bước giải:

Bài 5:

a.Xét $\Delta DBE,\Delta CBE$ có:

Chung $BE$

$\widehat{EBD}=\widehat{EBC}$

$BD=BC$

$\to \Delta EBD=\Delta EBC(c.g.c)$

b.Xét $\Delta FBD,\Delta FBC$ có:

Chung $FB$

$\widehat{FBD}=\widehat{FBC}$

$BD=BC$

$\to \Delta FBD=\Delta FBC(c.g.c)$

$\to FD=FC$

c.Từ b $\to \widehat{BFD}=\widehat{BFC}$

Mà $\widehat{BFD}+\widehat{BFC}=180^o$

$\to \widehat{BFD}=\widehat{BFC}=90^o$

$\to BF\perp CD$

Mà $Ah\perp CD$

$\to AH//BF$