Given triangle ABC such that angles B and C both measure 70 degrees, points E and F lie on sides AB and AC, respectively, such that angle ABF measures 30 degrees and angle ACE measures 50 degrees. Segments CE and BF intersect at point P. Find the measure of angle BAP.
Also, how do I generalize this problem with other angle measures? (i. e., when the angles of B and C and when the cevians necessarily split the angles are changed)
Yes, you can solve it with any angles. I'll only use the Sine Law. Let $CP$ intersect $AB$ at $K.~$ Let $AC=a.~$ Then $AB=a,~$
$$ BC=\dfrac{a\sin 40^{\circ}}{\sin70^{\circ}},\quad BP=\dfrac{a\sin 20^{\circ}\sin 40^{\circ}}{\sin 120^{\circ}\sin 70^{\circ}},\quad BK=\dfrac{a\sqrt{3}\sin 20^{\circ}\sin 40^{\circ}}{2\sin 120^{\circ}\sin 70^{\circ}}, $$
$$ KP=\dfrac{a\sin 20^{\circ}\sin 40^{\circ}}{2\sin 120^{\circ}\sin 70^{\circ}},\quad AK=\dfrac{a\sin(90^{\circ}-x)\sin 20^{\circ}\sin 20^{\circ}\sin 40^{\circ}}{2\sin x\sin 120^{\circ}\sin 70^{\circ}}. $$
Substitute those values of $AK$ and $BK$ into $AK+BK=a.~$ You'll get an equation in terms of $x$. The solution is $x=10^{\circ}.~$ See WolframAlpha.