In a triangle $ABC$ let $D$ be the midpoint of $BC$ . If $\angle ADB=45^\circ$ and $\angle ACD=30^\circ$ then find $\angle BAD$ In a triangle $\Delta ABC$ let $D$ be the midpoint of $BC$. If angle $\angle ADB=45^{\circ}$ and angle $\angle ACD=30^{\circ}$ then find angle $\angle BAD$. 
NOW this is a special case do we need a construction.

Drop perpendicular from $A$ on extended $BC$ intersecting at $E$ to form right triangle $\triangle AEC$.

Let $AC = b$
Then by basic trigonometry we get:
$EA=\frac{b}{2}$
$DC=EC-ED=\frac{\sqrt{3}-1}{2}b$
$EB=EC-BC=\frac{2-\sqrt{3}}{2}b$
$\angle EAB=\tan^{-1}\frac{EB}{EA}=\frac{\pi}{12}$
Therefore,
$$\angle BAD=\angle EAD-\angle EAB=\frac{\pi}{4}-\frac{\pi}{12}=\frac{\pi}{6}$$
