a) Xét $∆ABD$ và $∆HBD$ có:
$BH = BA \quad (gt)$
$\widehat{ABD}=\widehat{HBD}=\dfrac12\widehat{ABC}\quad (gt)$
$BD:$ cạnh chung
Do đó $∆ABD=∆HBD\, (c.g.c)$
$\to \widehat{BHD}=\widehat{BAD}=90^\circ$ (hai góc tương ứng)
$\to DH\perp BH$
$\to DH\perp BC$
b) Ta có:
$\widehat{B} =90^\circ -\widehat{C} = 90^\circ - 60^\circ = 30^\circ$
$\widehat{ABD}=\dfrac12\widehat{B}=\dfrac12\cdot 30^\circ = 15^\circ$
$∆ABD$ vuông tại $A$ có:
$\widehat{ABD}=15^\circ$
$\to \widehat{ADB}= 90^\circ - \widehat{ABD}$
$\to \widehat{ADB}=90^\circ - 15^\circ = 75^\circ$