a, Xét ΔABD và ΔEBD ta có:
BD chung
\(\widehat{ABD}\) = \(\widehat{EBD}\)
=> ΔABD = ΔEBD ( cạnh huyền - góc nhọn )
=> \(\widehat{ADB}\) = \(\widehat{EDB}\)
=> DB là phân giác \(\widehat{ADE}\)
b, \(\widehat{ABD}\) + \(\widehat{ADB}\) = \(\widehat{CKD}\) + \(\widehat{CDK}\) ( = 90 độ)
mà \(\widehat{ADB}\) = \(\widehat{CDK}\) ( hai góc đối đỉnh)
=> \(\widehat{ABD}\) = \(\widehat{CKD}\)
=> \(\widehat{CBD}\) = \(\widehat{CKD}\)
=> ΔBCK cân tại C
c, \(\widehat{ADB}\) = \(\widehat{BDE}\)
\(\widehat{ADH}\) = \(\widehat{EDC}\)
=> \(\widehat{ADB}\) + \(\widehat{ADH}\) = \(\widehat{BDE}\)+ \(\widehat{EDC}\)
=> \(\widehat{ADH}\) = \(\widehat{BDC}\)
Xét ΔBDH và ΔBDC ta có:
BD chung
\(\widehat{ABD}\) = \(\widehat{EBD}\)
\(\widehat{ADH}\) = \(\widehat{BDC}\)
=> ΔBDH = ΔBDC ( g-c-g)
=> BH = BC
mà BC = BK ( ΔBCK cân tại C)
=> BH = BK (đpcm)