a, OA = OB
OC = OD
=> AC = BD
Xét ΔACD và ΔBDC ta có:
CD chung
AC = BD
\(\widehat{ACD}\) = \(\widehat{BDC}\) (OC = OD)
=>ΔACD = ΔBDC (c-g-c)
=> AD = BC
b, ΔACD = ΔBDC
=> \(\widehat{ADC}\) = \(\widehat{BCD}\)
=> \(\widehat{BDC}\) - \(\widehat{ADC}\) = \(\widehat{ACD}\) - \(\widehat{BCD}\)
=> \(\widehat{ACE}\) = \(\widehat{BDE}\)
ΔACD = ΔBDC => \(\widehat{CAD}\) = \(\widehat{DBC}\)
Xét ΔAEC và ΔBED ta có:
\(\widehat{CAD}\) = \(\widehat{DBC}\)
\(\widehat{ACE}\) = \(\widehat{BDE}\)
AC = BD
=> ΔAEC = ΔBED ( g-c-g) (đpcm)