In $\triangle ABC$, $D$ lies on $BC$, $\angle ADB=60$, $\angle ACB=45$ and $BD=2CD$. Find $\angle ABC$ The question is as the title states. In the following figure, with some given angles and two sides, the goal is to find the measure of $\angle ABC$. This problem is inspired by one which appeared in a local Math contest in Turkey. I'll post my approach to this below as an answer! Please show your own methods of solving this! 
 A: Here's my approach. I'll post a brief explanation below!

1.) Draw a perpendicular from $D$ onto line segment $AD$ such that it meets at point $E$. Join point $E$ with $C$ via $EC$. Notice that $\triangle BED$ is a $30-60-90$ triangle, therefore, $ED=DC=a$, and $\angle EBD=\angle DEC=\angle DCE=30$. We also know that $BE=EC=a\sqrt3$.
2.) Note than $\angle DAC=\angle ACE=15$, therefore $BE=EC=AE=a\sqrt3$. However, this proves that $\triangle AEB$ is an isosceles right angle triangle, therefore $\angle ABE=45$, and thus, $\angle ABC=45+30=75$
A: Alternative approach:
From the original figure, in the original posting, construct the perpendicular line segment from point $A$ to the line segment $\overline{BC}$.  Assume that the perpendicular line segment intersects $\overline{BC}$ at the point $G$.

*

*Let $r$ denote the length of $\overline{AG}.$

*Let $s$ denote the length of $\overline{BG}.$

*Let $t$ denote the length of $\overline{DG}.$
The problems asks that $\angle ABC$ be computed.
Then:

*

*$\displaystyle \frac{r}{a+t} = \tan(45^\circ) = 1.$

*$\displaystyle \frac{r}{t} = \tan(60^\circ) = \sqrt{3}.$

*$\displaystyle \frac{r}{s} = \tan(\angle ABC).$

*$(s + t) = 2a.$
Then,
$$\frac{a+t}{t} = \frac{\frac{r}{t}}{\frac{r}{a + t}} = \sqrt{3} \implies $$
$$a + t = t\sqrt{3} \implies a = t\left[~-1 + \sqrt{3}~\right] \implies $$
$$s = 2a - t = t\left[~-3 + 2\sqrt{3}~\right] \implies $$
$$\tan\left(\angle ABC\right) = \frac{r}{s} = \frac{r}{t\left[~-3 + 2\sqrt{3}~\right]}$$
$$= \frac{r}{t} \times \frac{1}{\left[~-3 + 2\sqrt{3}~\right]}$$
$$= \frac{\sqrt{3}}{\left[~-3 + 2\sqrt{3}~\right]}.$$
Therefore,
$$\angle ABC = \text{Arctan}\left\{ ~ \frac{\sqrt{3}}{\left[~-3 + 2\sqrt{3}~\right]} ~\right\} \implies \angle ABC = 75^\circ.$$
A: 
Draw $AE\perp BC$ and $AF$ such that $\angle CAF=15^o$
Let $DE=b$ then $BE=2a-b$, $AE=EC=b\sqrt3$, $EF=3b$ and $AF=2b\sqrt{3}$
$AE=EC=b\sqrt3=a+b$ then $a=b\sqrt3 - b$
$BF=AE+EF=(2a-b)+3b=2(a+b)=2(b\sqrt3-b+b)=2b\sqrt{3}$
Since $BF=BA$ and $\angle AFB=30^o$ then $\angle B=75^o$
A: Let $\angle ABC = x $, then applying the law of sines on $\triangle ADC $:
$\dfrac{AD}{\sin 45^\circ} = \dfrac{a}{\sin 15^\circ} $
Applying the law of sines on $\triangle ADB$:
$\dfrac{AD}{\sin x} = \dfrac{2 a}{\sin(120^\circ - x) } $
Dividing the two equation to eliminate $AD$ and $a$
$ \dfrac{ \sin x }{ \sin 45^\circ } = \dfrac{1}{2} \dfrac{ \sin(120^\circ - x) }{ \sin 15^\circ } $
By cross-multiplying,
$ 2 \sin x \sin 15^\circ = \sin 45^\circ \bigg( \sin 120^\circ \cos x - \cos 120^\circ \sin x \bigg) $
Dividing through by $\cos x $
$ 2 \tan x \sin 15^\circ = \sin 45^\circ \bigg( \sin 120^\circ - \cos 120^\circ \tan x \bigg)$
Hence,
$ \tan x = \dfrac{ \sin 45^\circ \sin 120^\circ }{ 2 \sin 15^\circ + \sin 45^\circ \cos 120^\circ } $
It follows that
$ x = 75^\circ $
