Hard about Concurrence As shown in the figure, note that P is a point of concurrency. 
How we can prove it geometrically ?
Any hints would be appreciated.

 A: We may draw the two triangles assuming we know all angles except for the two at the top, and may as well take $AC=1.$ Then using the law of sines we have $AP=\sin 62/\sin106$ and $AB=\sin70/\sin52.$ Now let $u$ be the angle marked $50$, i.e. angle $ABP$, and $v$ be the angle $BPA$. Note that in $\Delta APB$ we already know the angle $PAB$ is $46$. So the unknown angles $u,v$ satisfy two equations: $u+v=180-46=134$ and the equation using the law of sines on $\Delta APB$, namely
$$\frac{\sin u}{(\sin 62/\sin 106)}=\frac{\sin v}{(\sin 70 / \sin 52)}.\tag{1}$$
When done numerically, this gives $u=50$. (If $u=50$ is known the rest follows.)
It seems clear the two equations for $u,v$ have a unique solution, so perhaps one way to finish would be to plug in $u=50,\ v=84$ and prove somehow using trig identities that equation $(1)$ holds for those angles. I tried that a bit without success, but there may be a way.
added. the exact equality $(1)$ (with its $u,v$ as $50,84$) can indeed be shown using trig. One cross-multiplies and gets two triple products of sines which one wants to be equal. Then apply the product-to sum formulas to both sides until each is a sum of only cosines, and the terms all match.
Another approach to the problem can be based on Ceva's theorem. We can get all six ratios in which the points where the rays from the vertices intersect opposite sides cut their respective sides, using the law of sines. Ceva's theorem then says the rays meet if and only if the product of these ratios is 1, and what we wind up with is similar to the approach above, namely we want again to show that two certain products each of three sines are equal. And again by product to sum formulas it turns out they are. The approach using Ceva's theorem is in a way better, as it doesn't assume at the outset that there is a common point to the three rays. It just says that, if the ratio product is 1, then the rays concur. The ratios could be computed from any initial data as in the diagram (i.e. the various marked angles could be changed), and Ceva's theorem would give an answer as to whether the rays meet, provided one could validly show whether or not the sine ratios multiply to 1.
A: Invoke the trigonometric form of Ceva's Theorem:
$$
\frac{\sin\angle ABP}{\sin\angle CBP}\cdot
\frac{\sin\angle BCP}{\sin\angle ACP}\cdot
\frac{\sin\angle CAP}{\sin\angle BAP} \;=\;
\frac{\sin 50^\circ}{\sin 2^\circ}\cdot
\frac{\sin 8^\circ}{\sin 62^\circ}\cdot
\frac{\sin 12^\circ}{\sin 46^\circ} \quad \stackrel{?}{=}\quad 1
$$
(In)Equality indicates that the concurrency is (in)valid.
Mathematica numerically evaluates the fraction of sines as $1$, but I'm not seeing how to verify this symbolically.
A: You can calculate $AP$ in two ways, from triangle $APB$ - where the angle opposite side $AB$ is $84^{\circ}$, and from triangle $APC$ where the angle at $P$ is $106^{\circ}$. Also let $R$ be the circumradius of $\triangle ABC$
So $\cfrac {AB}{\sin 84^{\circ}}=\cfrac {AP}{\sin 50^{\circ}}$ and $\cfrac {AC}{\sin 106^{\circ}}=\cfrac {AP}{\sin 62^{\circ}}$
So $AP=\cfrac {AB\sin 50^{\circ}}{\sin 84^{\circ}}=\cfrac {AC\sin 62^{\circ}}{\sin 106^{\circ}}$
We apply the "long form" of the sine rule in $\triangle ABC$ to give:
$2R=\cfrac {AB}{\sin 70^{\circ}}=\cfrac {AC}{\sin 52^{\circ}}$ so that$$AP=\cfrac {2R\sin 70^{\circ}\sin 50^{\circ}}{\sin 84^{\circ}}=\cfrac {2R\sin 52^{\circ}\sin 62^{\circ}}{\sin 106^{\circ}}$$ We equate the two fractions to find the condition:$$\frac {{\sin 70^{\circ}\sin 50^{\circ}\sin 106^{\circ}}}{\sin 52^{\circ}\sin 62^{\circ}\sin 84^{\circ}}=1$$
