a: Xét ΔBAD và ΔBHD có
BA=BH
\(\widehat{ABD}=\widehat{HBD}\)
BD chung
Do đó: ΔBAD=ΔBHD
b: ta có: ΔBAD=ΔBHD
nên \(\widehat{BAD}=\widehat{BHD}=90^0\)
=>DH\(\perp\)BC
c: \(\widehat{ABC}=90^0-30^0=15^0\)
=>\(\widehat{CBD}=15^0\)
\(\widehat{BDC}=180^0-75^0=105^0\)