a, \(\widehat{ACB}\) = \(\widehat{BAC}\) - \(\widehat{ABC}\) = 90\(^{\circ}\) - 60\(^{\circ}\) = 30\(^{\circ}\)
\(\widehat{EBA}\)= \(\widehat{EBC}\) = \(\frac{\widehat{ABC}}{2}\) = \(\frac{\widehat{60^{\circ}}}{2}\) = 30\(^{\circ}\)
b, Ta có: \(\widehat{ACB}\) = \(\widehat{EBC}\) (= 30\(^{\circ}\))
=> ΔCEB cân tại E
EC = EB
c, Xét ΔABC và ΔDCB ta có:
\(\widehat{ACB}\) = \(\widehat{EBC}\)
BC chung
=> ΔABC = ΔDCB ( cạnh huyền - góc nhọn)
AC = DB