Xét tứ giác AEID có
\(\widehat{AEI}+\widehat{ADI}+\widehat{EAD}+\widehat{EID}=360^0\)
=>\(\widehat{EAD}+\widehat{EID}+90^0+90^0=360^0\)
=>\(\widehat{EAD}+\widehat{EID}=360^0-180^0=180^0\)
mà \(\widehat{EID}=\widehat{BIC}\)(hai góc đối đỉnh)
nên \(\widehat{EAD}+\widehat{BIC}=180^0\)
=>góc BIC bù với góc BAC