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.
 A: Let's denote $\angle BAD$ as $\alpha$. We have that $\angle ADB=180-(45+\alpha)$. If $\angle ADB=45$ and $\angle ACD=30$, then $\angle ADC=180-45$ and $\angle CDA=180-30-(180-45)=15$. Now, from the sine theorem:
$$ \frac{|CD|}{\sin(15)}=\frac{|AC|}{\sin(180-45)}$$
$$ \frac{|AC|}{\sin(180-(45+\alpha))}=\frac{|BD|}{\sin(\alpha)}=\frac{|CD|}{\sin(\alpha)}$$
From the first one, we have that: 
$$ |AC|=\frac{\sin(180-45)}{\sin(15)}|CD|$$
Plugging this into the second one gives us:
$$ \frac{\sin(180-45)}{\sin(15)}|CD|\frac{1}{\sin(180-(45+\alpha))}=\frac{|CD|}{\sin(\alpha)} $$
$$ \frac{\sin(15)\sin(180-45)}{\sin(180-(45+\alpha))}=\frac{1}{\sin(\alpha)} $$
$$ \sin(\alpha)=\frac{\sin(180-(45+\alpha))}{\sin(15)\sin(180-45)}=\frac{\sin(45+\alpha)}{\sin(15)\sin(180-45)} $$
Can you take it from here? 
A: One can use a construction if one does not want to use trigonometry but only very simple but tricky methods. 
Let point $B'$ be such that triangle $AB'C$ is equilateral and $B$ and $B'$ lie on the same side of line $AC.$ Let point $C'$ be such that line $AC'$ is orthogonal to $B'C$ and $AC' = AC = AB' = B'C$. By construction, $AC'$ is the orthgognal bisector of $B'C$ so $$B'C' = CC'\,\,\, \text{ and } \,\,\, \angle \, CAC' = B'AC' = 30^{\circ}$$ Therefore, triangles $CAC'$ and $B'AC'$ are congruent isosceles triangles with $30^{\circ}$ internal angle at vertex $A$ (because $AC'$ is the angle bisector of equlateral triangle $AB'C$).

Now, by assumption, $\angle \, DAC = \angle \, ADB - \angle \, ACD = 45^{\circ} - 30^{\circ} = 15^{\circ}$. Hence, line $AD$ is the angle bisector in isosceles triagnle $CAC'$ through vertex $A$ and so $AD$ is also orthogonal bisectror of $CC'$ which means that $DC = DC'$. A direct angle chasing shows that $\angle \, CDC' = 90^{\circ}$. However, by assumption, $BD = DC = DC'$ so $$BC ' = CC' = B'C' \,\,\, \text{ and } \,\,\, \angle \, BC'C = 90^{\circ}$$ Hence triangle $BB'C'$ is isosceles and after a direct angle chasing, one sees that $\angle \, BC'B' = 60^{\circ}$. Thus $BB'C'$ is an equilateral triangle. This means that triangles $ABC'$ and $ABB'$ are congruent, because $BB' = BC'$, segment $AB$ is common and $AB'=AC'$ by construction. Therefore, $$\angle \, BAB' = \angle \, BAC' = \frac{1}{2} \, \angle \, B'AC' = \frac{1}{2} \, 30^{\circ} = 15^{\circ}$$ We can conclude that $$\angle \, BAD = \angle \, B'AC - \big(\angle \, BAB' + \angle DAC \big) = 60^{\circ} - (15^{\circ} + 15^{\circ}) = 30^{\circ}$$
A: This can be solved very easily using a simple geometry construction:

1.) Draw segment $DE$ where point $E$ lies on $AC$ such that $\angle DAC=\angle ADE=15$. This means that $\angle DEC=30$, therefore $AE=ED=BD=DC$. Now, join $B$ and $E$ via $BE$. Notice that $\angle BDE=60$ and $BD=DE$, this proves that $\triangle BDE$ is equilateral, and therefore, $AE=ED=BD=DC=BE$.
2.) Lastly, notice that $\angle BEC=\angle BEA=90$, and that $AE=BE$, this means $\triangle BAE$ is an isosceles right triangle, and $\angle BAE=45$. Therefore $\angle BAD=45-15=30$
A: 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,
$$\angle BAD=\angle EAD-\angle EAB=\frac{\pi}{4}-\frac{\pi}{12}=\frac{\pi}{6}$$
