Geometry

BTW I need this quick! :)

Use Pythagoras theorem:

$AC=\sqrt{25^2-7^2}=24$

$CD=DA=12$(mid point)

Use Pythagoras theorem again:

$BD=\sqrt{7^2+12^2}=\sqrt{193}$

Find the sin of angle BCD:

$\sin \angle BCD =\sin \angle BCA = \dfrac{7}{25}$(angle BCD and angle BCA refers to the same angle)

Use law of sines on triangle BCD:

$\dfrac{BD}{\sin\angle BCD}=\dfrac{BC}{\sin\angle BDC}\\ \dfrac{\sqrt{193}}{\frac{7}{25}}=\dfrac{25}{\sin\angle BDC}\\ \sin\angle BDC = \dfrac{25\times \frac{7}{25}}{\sqrt{193}}=\dfrac{7}{\sqrt{193}}$

Construct a right-angled triangle with

legs = 7, -12(<-- negative because the angle is in the 2nd quadrant), hypotenuse=sqrt193

$\tan \angle BDC =-\dfrac{7}{12}$

Quicker way:

angle BDC = 90 deg + angle DBA

$\boxed{\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a \cos b+\cos a \sin b}{\cos a \cos b - \sin a \sin b}=\dfrac{\tan a + \tan b}{1-\tan a \tan b}}$

But because tan 90 deg does not have a meaning, we use the old definition:

$\tan 90^{\circ}=\dfrac{1}{0}$

So 

$\dfrac{1}{\tan 90^{\circ}}=0$

$\tan \angle BDC \\= \tan (90^{\circ}+\angle DBA)\\ =\dfrac{\tan 90^{\circ}+\tan{\angle DBA}}{1-\tan90^{\circ}\tan\angle DBA}\\ =\dfrac{1+\frac{\tan \angle DBA}{\tan 90^{\circ}}}{\dfrac{1}{\tan 90^{\circ}}-\tan \angle DBA}\\ =\dfrac{1+0\tan\angle DBA}{0-\tan\angle DBA}\\ =-\dfrac{1}{\tan\angle DBA}$

AC = 24 using Pyth. thm. <-- I am lazy in typing

AD = 24/2 = 12

So tan angle DBA = 12/7

So tan angle BDC = -1/tan angle DBA = -7/12 <-- done :)

Since BDA and BDC are supplemental, tan( BDA)   =  - tan (BDC) →  - tan (BDA)  = tan (BDC)

Using right angle trigonometry,  tan (BDA)  =  7/12

So  -tan(BDA)  =  -7/12  = tan (BDC)