One way is to do it numerically for example with wolframalpha.
use Li Li solution
$$\frac{AD^2+CD^2-AC^2}{2AD\cdot CD}=\cos(\angle ADC)$$
$$\frac{BD^2+CD^2-BC^2}{2BD\cdot CD}=\cos(\angle BDC)$$
$$\frac{AD^2+BD^2-AB^2}{2AD\cdot BD}=\cos(\angle ADB)$$
like here, it's a example of 2 second degree equations
Yes, there is other method: analytic geometry.
Points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are collinear if and only if
$$\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\\\end{vmatrix}=0 $$
Using it we can easily solve the question:
For instance,we can put the origin of oblique cartesian axes at point B such that B, A, E lie in y-axis and B, C, F lie in x-axis. Then B has coordinates $B=(0,0)$ and without loss of generality the coordinates of A, C, D, E, F can be $A=(0,2a),C=(2c,0),D=(2d_1,2d_2),E=(0,2e),F=(2f,0)$.
As points C, D, E are collinear, we get
$$\begin{vmatrix}2c & 0 & 1\\2d_1 & 2d_2 & 1\\0 & 2e & 1\\\end{vmatrix}=0, $$
$$\begin{vmatrix}c & 0 & 1\\d_1 & d_2 & 1\\0 & e & 1\\\end{vmatrix}=0. $$
On the other hand, points A, D, F are also collinear, then
$$\begin{vmatrix}0 & 2a & 1\\2d_1 & 2d_2 & 1\\2f & 0 & 1\\\end{vmatrix}=0, $$
$$\begin{vmatrix}0 & a & 1\\d_1 & d_2 & 1\\f & 0 & 1\\\end{vmatrix}=0. $$
Summing both determinants, we get
$$c(d_2-e)+a(f-d_1)+ 1(ed_1-fd_2)=0,$$
$$\begin{vmatrix}c & a & 1\\d_1 & d_2 & 1\\f & e & 1\\\end{vmatrix}=0. $$.
Therefore the points $(c,a),(d_1,d_2), (f,e)$, aka midpoints of AC, BD, EF, are collinear.
QED.
Best Answer
Here's my solution:
With $AB$ as a side, construct equilateral triangle $\triangle AEB$, such that $AB=EB=AB$. Notice that $\angle ADB=150^\circ$, this means that $E$ is the circumcenter of $\triangle ADB$, this implies that $AE=EB=ED=AB$. This also means that $\angle DEB=24^\circ$ and $\angle DEA=36^\circ$.
We know that $\angle DBE=78^\circ$, which means that $\angle BDE=78^\circ$ as well. Lastly, notice that $\angle BDC=102^\circ$, which means that points $E$, $D$ and $C$ are collinear. Since $\angle BEC=\angle BCE=24^\circ$, we know that $BE=BC$ which proves that $AB=BC$.