Giải thích các bước giải:

a.Xét $\Delta FAE, \Delta FCH$ có:

$FA=FC$

$\widehat{AFE}=\widehat{CFH}$

$FE=FH$

$\to \Delta FAE=\Delta FCH(c.g.c)$

$\to AE=HC$

b.Xét $\Delta FAH, \Delta FCE$ có:

$FA=FC$

$\widehat{AFH}=\widehat{EFC}$

$FH=FE$

$\to \Delta FAH=\Delta FCE(c.g.c)$

$\to \widehat{FAH}=\widehat{FCE}$

$\to AH//CE$

c.Xét $\Delta FAI,\Delta FCQ$ có:

$\hat I=\hat Q(=90^o)$

$FA=FC$

$\widehat{AFI}=\widehat{CFQ}$

$\to \Delta FAI=\Delta FCQ$(cạnh huyền-góc nhọn)

$\to FI=FQ$

Xét $\Delta FAQ, \Delta FCI$ có:

$FA=FC$

$\widehat{AFQ}=\widehat{CFI}$

$FQ=FI$

$\to \Delta FAQ=\Delta FCI(c.g.c)$

$\to \widehat{FAQ}=\widehat{FCI}$

$\to AQ//CI$

d.Từ a $\to \widehat{EAF}=\widehat{FCH}\to HC//AB$

Xét $\Delta CEH, \Delta CEB$ có:

Chung $CE$

$\widehat{CEB}=\widehat{ECH}$ vì $HC//AB$

$CH=BE(=AE)$

$\to \Delta CEH=\Delta ECB(c.g.c)$

$\to BC=EH=2EF$