Use Pythagoras theorem:
AC=√252−72=24
CD=DA=12(mid point)
Use Pythagoras theorem again:
BD=√72+122=√193
Find the sin of angle BCD:
sin∠BCD=sin∠BCA=725(angle BCD and angle BCA refers to the same angle)
Use law of sines on triangle BCD:
BDsin∠BCD=BCsin∠BDC√193725=25sin∠BDCsin∠BDC=25×725√193=7√193
Construct a right-angled triangle with
legs = 7, -12(<-- negative because the angle is in the 2nd quadrant), hypotenuse=sqrt193
tan∠BDC=−712
.