Bài 3:
c) $\triangle$ABC cân
$\Rightarrow$ $\widehat{B}$ = $\widehat{C}$
$\widehat{A}$=4$\widehat{B}$ $\rightarrow$$\widehat{A}$+$\widehat{B}$+$\widehat{C}$=$180^o$
$\leftrightarrow$4$\widehat{B}$+$\widehat{B}$+$\widehat{B}$=$180^o$
$\leftrightarrow$ $\widehat{B}$ = $30^o$=$\widehat{C}$
$\leftrightarrow$$\widehat{A}$=$120^o$
Bài 4:
a) $\triangle$ABC và $\triangle$FBE có:
AB=FB ; $\widehat{B1}$=$\widehat{B2}$ ; BE chung
$\Rightarrow$$\triangle$ABC = $\triangle$FBE (c.g.c)
$\Rightarrow$ AE=EF ; $\widehat{EFB}$=$\widehat{A}$=$90^o$
b) $\triangle$AEK và $\triangle$FEC có:
$\widehat{A}$=$\widehat{E}$=$90^o$; AE=FE; $\widehat{E1}$=$\widehat{E2}$ ( đối đỉnh)
$\Rightarrow$ $\triangle$AEK = $\triangle$FEC (g.c.g)
$\Rightarrow$ EK = EC
c) $\triangle$ABF cân tại B có BE là phân giác
$\Rightarrow$ BE $\bot$ AF
