a) Vì H là trọng tâm của tam giác ABD đều nên :
\(HA=HB\)
\(\widehat{HAB}=\widehat{HBA}=\dfrac{1}{2}\widehat{DAB}=30^0\)
Xét \(\bigtriangleup{HBI} =\bigtriangleup{KCI}\) (c.g.c)
\(=> HB =CK\)
\(=>AH=CK\)
b, Ta có :
\(\widehat{HAE}= \widehat{HAB} + \widehat{BAC}+\widehat{CAE} \)
\(=30°+\widehat{BAC}+60°=90°+\widehat{BAC} (1)\)
Lại có: \(\widehat{KCE}= 360° - \widehat{ACE} - \widehat{ACB} - \widehat{BCK}\)
\(=> \widehat{KCE}= 360°- 60°- \widehat{ACB}- \widehat{HBI}\)
\(=> \widehat{KCE}= 300° - \widehat{ACB}- \widehat{HBA}- \widehat{ABC}\)
\(=> \widehat{KCE}= 300°- (\widehat{ABC}+\widehat{ACB})-30°\)
\(=> \widehat{KCE}= 270°- ( 180°- \widehat{BAC})\)
\(=> \widehat{KCE}= 90°+ \widehat{BAC} (2)\)
Từ (1) và (2) => \(\widehat{KCE} =\widehat{HAE}\)
=> \(\bigtriangleup{AHE}=\bigtriangleup{CKE}\)
c, Ta có:
\(\widehat{AEH}+ \widehat{HEC}=\widehat{AEC}=60°\)
Mà \(\widehat{AEH}= \widehat{CEK}( \bigtriangleup{ AHE} = \bigtriangleup{ CKE})\)
=> \(\widehat{CEK} +\widehat{HEC}= 60° => \widehat{HEK} =60°(3)\)
Mặt khác: HE= KE(\(\bigtriangleup{AHE}=\bigtriangleup{CKE}\))
=> \(\bigtriangleup\) HEK cân tại E (4)
Từ (3) và (4) =>\(\bigtriangleup\) HEK đều
