a, \(\widehat{ABC}\) = \(\widehat{BCA}\) - \(\widehat{CAB}\) = 90\(^{\circ}\) - 60\(^{\circ}\) = 30\(^{\circ}\)
\(\widehat{CAE}\) = \(\widehat{EAB}\) = \(\frac{\widehat{CAB}}{2}\)= \(\frac{60^{\circ}}{2}\) = 30\(^{\circ}\)
=> \(\widehat{AEB}\) = \(\widehat{CAE}\) + \(\widehat{ECA}\) = 90\(^{\circ}\) + 30 \(^{\circ}\)= 120\(^{\circ}\)
\(\widehat{AEC}\) = \(\widehat{ECA}\) - \(\widehat{CAE}\) = 90\(^{\circ}\) - 30\(^{\circ}\) = 60\(^{\circ}\)
b, Xét ΔECA và ΔEKA ta có:
EA chung
\(\widehat{CAE}\) = \(\widehat{KAE}\) (= 30\(^{\circ}\))
=> ΔECA = ΔEKA ( cạnh huyền - góc nhọn)
=> EC = EK
c, ΔBEA có \(\widehat{KAE}\) = \(\widehat{EBA}\) (= 30\(^{\circ}\))
=> ΔBEA cân tại E
=> EB = EA
Xét ΔDBE và ΔCAE ta có:
EB = EA
\(\widehat{DEB}\) = \(\widehat{CEA}\) ( đối đỉnh)
=> ΔDBE = ΔCAE ( cạnh huyền - góc nhọn)
=> BD = CA
d, ΔDBE = ΔCAE
=> \(\widehat{DBC}\) = \(\widehat{CAE}\) = 30\(^{\circ}\)
=> \(\widehat{IBA}\) = \(\widehat{DBC}\) + \(\widehat{CBA}\) = 30\(^{\circ}\) + 30\(^{\circ}\) = 60\(^{\circ}\)
Xét ΔIBA có \(\widehat{IBA}\) = \(\widehat{IAB}\) = 60\(^{\circ}\)
=> ΔIBA đều