Câu 4.6:
a. Xét ΔABD và ΔCBD, ta có:
DA=DC(gt); BD chung; BA=BC
=> ΔABD=ΔCBD(c.c.c)
b)Ta có $\widehat{DAB} $= $\widehat{DCB} $ = $90^{\circ}$ (hai góc tương ứng)
Mà $\widehat{DCB} $+ $\widehat{CDB} $ + $\widehat{DBC} $= $180^{\circ}$ => $\widehat{DBC} $ = $180^{\circ}$- ($\widehat{DCB} $+ $\widehat{CDB} $ ) = $60^{\circ}$
Mà ΔABD=ΔCBD =>
nên $\widehat{ABD}$ = $\widehat{CBD}$( 2 góc tương ứng)
=>$\widehat{ABD}$ = $60^{\circ}$ => $\widehat{ABC}$= $60^{\circ}$+ $60^{\circ}$= $120^{\circ}$