First of all, the question is printed incorrectly, and the conclusion should be ∠PBA=∠ACB (not ∠PBA=∠PCA).

PC and AE meet at Q.

Aq/AE = s △ baq/s △ BAE = s △ baq/s △ ADC = s △ baq/s △ APC (because of parallelism)

S△BAQ=AB*AQ*sin∠BAE/2

S△APC=AC*AP*sin∠PAC/2

S△BAQ/S△APC=AB*AQ/(AC*AP)

AB/AP=AC/AE similar

This problem area method is the simplest (because BD=CE, PD//AE conditions are not easy to convert).

Parallel axiom

Not as obvious as other axioms. Many geometricians tried to prove this axiom with other axioms, but all failed. /kloc-in the 9th century, by constructing non-Euclidean geometry, it was proved that parallel axioms could not be proved (if parallel axioms were removed from the above axiom system, a more general geometry, that is, absolute geometry, could be obtained).

On the other hand, the five axioms (postulates) of Euclid's geometry are incomplete. For example, all theorems in this geometry: any line segment is part of a triangle. He constructed it in the usual way: taking the line segment as the radius, taking the two endpoints of the line segment as the center respectively, and taking the intersection of two circles as the third vertex of the triangle.