If $P$ is the midpoint of side $BC$ of triangle $ABC$, and if $AB<AC$, prove that $\angle PAC<\angle BAP$

Question

This is a question from College Geometry by Howard Eves - Section 1.1 - Problem 10

If $P$ is the midpoint of side $BC$ of triangle $ABC$, and if $AB<AC$, prove that $\angle PAC<\angle BAP$

I am not sure how to start this proof, I know midpoint is defined as:

$$P=\frac{1}{2}\left(B+C\right)$$

I did a graph, but I am not sure how to proceed.

Answer 1

Translate triangle $ABP$ to $A'PC$, to obtain parallelogram $AA'CP$ (figure below): in triangle $A'OC$ you have $OC>OA'$, whence $\angle OA'C>\angle OCA'$.