Ta có: \(\widehat{BAH}=\widehat{C}\)
mà \(\widehat{C}< \widehat{B}\)
nên \(\widehat{BAH}< \widehat{B}\)
=>BH<AH(1)
Ta có: \(\widehat{HAC}=\widehat{B}\)
mà \(\widehat{B}>\widehat{C}\)
nên \(\widehat{HAC}>\widehat{C}\)
=>AH<HC(2)
Từ (1)và (2) suy ra BH<AH<HC