Giải thích các bước giải:
a, $\widehat{BAE}$ = $\widehat{BAC}$ + $60^o$
$\widehat{DAC}$ = $\widehat{BAC}$ + $60^o$
⇒ $\widehat{BAE}$ = $\widehat{DAC}$ (đpcm)
b, ΔABE và ΔADC có:
AB = AD; AE = AC (gt);
$\widehat{BAE}$ = $\widehat{DAC}$;
⇒ ΔABE = ΔADC (c.g.c)
⇒ BE = DC (đpcm)
c, ΔABE = ΔADC (c.g.c) ⇒ $\widehat{ABE}$ = $\widehat{ADC}$
⇒ $\widehat{ABE}$ + $\widehat{ACD}$ = $\widehat{ACD}$ + $\widehat{ADC}$ = $180^o - \widehat{DAC}$ = $180^o - 120^o = 60^o$
⇒ $\widehat{BKC}$ = $180^o - (\widehat{KBC} + \widehat{KCB})$
= $180^o - [180^o - (\widehat{BAC} + \widehat{ABE} + \widehat{ACD})]$
= $60^o + 60^o$ = $120^o$