Given a triangle whose vertex $A$ is bisecting the base (median), I have to find the angle $X$.
First Approach
From $\Delta AHD$ we can get $AH = HD$. And it will make $45^\circ$ at vertex $A^*$. From $\Delta ABH$, let's Consider angle at $B$ is $\theta$: $\tan \theta = \frac{AH}{BH}$.
From here, I am not able to think forward.
Assumption: If somehow I can prove that $BH = HD$, then $\Delta ABH$ will also be an isosceles triangle. Therefore total angle at vertex $A$ will be $45^\circ + 45^\circ = 90^\circ$.
Possible Second Approach
Assuming angle at $B$ is $Y$, from $\Delta ABD$,
$X + Y + 45^\circ = 180^\circ\\X+Y=135^\circ \qquad(I).$
From $\Delta ABC$: $(X + 15) + Y + 30^\circ = 180^\circ\\X+Y=135^\circ \qquad(II)$
So, from above two equations I am not able to find $X$ and $Y$ (Since, Both are same)
If I can get another linear equation in terms of $X$ and $Y$. By solving those equation, I can get angle at '$A$'
Best Answer
$\require{cancel}$
Using completely trigonometry
Suppose $AC=q$, let us solve for all the sides in terms of $q$: $$\frac{\sin(135^\circ)}{q}=\frac{\sin(30^\circ)}{AD}\implies AD=\frac{q}{\sqrt2}$$ $$\frac{\sin(15^\circ)}{CD}=\frac{\sin(135^\circ)}{q}\implies\,CD=\frac{q(\sqrt3-1)}{2}$$ Since $CD=BD$, then $BC=q(\sqrt3-1)$. Then we need to find $AB$: $$AB^2=q^2+\left(q\left(\sqrt3-1\right)\right)^2-2q^2(\sqrt3-1)\cos(30^\circ)\implies AB=q\sqrt{2-\sqrt{3}}$$ Then we just need to solve for $x$ in: $$\frac{\sin x}{\frac{q(\sqrt3-1)}2}=\frac{\sin 45^\circ}{q\sqrt{2-\sqrt3}}\implies \sin x=\frac{\sqrt3-1}{2\sqrt{4-2\sqrt3}}$$ The nested radical in the bottom looks ugly, but remember that the radical $\sqrt{a\pm b\sqrt{c}}=\sqrt d\pm\sqrt e$ if and only if $\sqrt{a^2- b^2c}\,\,$ is rational. Therefore, we have: $$\sin x=\frac{\cancel{\sqrt3-1}}{2\cancel{(\sqrt3-1)}}=\frac12 \implies$$ $$\bbox[10px, border: 2px black solid]{\therefore x=\arcsin{\frac12}=\frac\pi6=30^\circ}$$