a: Xét ΔBAD và ΔBHD có
BA=BH
\(\widehat{ABD}=\widehat{HBD}\)
BD chung
Do đo: ΔBAD=ΔBHD
b: ta có: ΔBAD=ΔBHD
nên \(\widehat{BAD}=\widehat{BHD}=90^0\)
hay DH\(\perp\)BC
c: \(\widehat{ABC}=90^0-60^0=30^0\)
=>\(\widehat{DBC}=15^0\)
\(\widehat{BDC}=180^0-60^0-15^0=105^0\)