In triangle $ABC$, $AB = 5$, $BC = 4$, and $CA = 3$. Point $P$ is randomly selected inside triangle $ABC$. What is the probability that $P$

I believe this might be correct, Mellie.......look at the pic.....

 Note....{DG ll CE   and GE ll DC }

Notice, that in triangle ABC, that if we pick any point "P" on DG, then triangle APD will be congruent to triangle CPD by SAS. Thus CP will be  equal to AP. But, for any point "P" that we select inside triangle CDG, then point "P" will be further from A - and certainly further from B - than "P" is from C. This is so because angle CAP will be less than angle ACP. And in any triangle, the greater angle is opposite the greater side. Thus, AP will be greater than CP.

And for the same reasons, any point "P" on GE, will make triangles BPE and CPE congruent. But, any point "P" inside triangle CEG will be further from B - and certainly further from A - than "P" is from C.

Thus....all points in rectangle CDGE will be futher from A or B than they are from C. And the area of this rectangle is just (2)(1.5)  = 3 sq units.

And the area of triangle ABC  = (1/2)(4)(3)  = 6 sq units.

So the probability that some point "P" is clser to C tham to either A or B  =  3 / 6   = 1 / 2.

that looks interesting Chris :)