a) Ta có \(\widehat{HAF}+\widehat{FAB}+\widehat{DAB}+\widehat{DAH}=360^o\)
mà \(\widehat{FAB}=\widehat{DAH}=90^o=>\widehat{HAF}+\widehat{DAB}=180^o\)
\(\widehat{ADC}+\widehat{DAB}=180^o=>\widehat{HAF}=\widehat{ADC}\)
Xét ΔHAF và ΔADC có HA = HD ; \(\widehat{HAF}=\widehat{ADC};AF=DC\)
=> ΔHAF = ΔADC (c.g.c) => FH = CA