What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP$? 
A point $P$ is chosen at random in the interior of the equilateral
  triangle $ABC$.What is the probability that $\triangle ABP$ has a
  greater area than each of $\triangle ACP$ and $\triangle BCP$?


Since the three triangles $ABP,ACP$ and $BCP$ have equal bases,their areas are proportional to their length of altitudes.
But i dont know how to solve further and find the required probability.Please help me.Thanks. 

Let distance from P to AB, AC and BC to be $L_{ab}$, $L_{ac}$ and $L_{bc}$, respectively.
Then $L_{ac}+L_{ac}+L_{bc}$ is constant and equal to the distance from A to BC (let's call this L), because sum of areas of three triangles ABP, ACP and BCP should be equal to the area of ABC.
So the condition becomes
$$L_{ab} > L_{ac}$$
$$L_{ab} > L_{bc}$$
So, P should exist within the region defined by


*

*Center of ABC

*C

*Mid point of A-C

*Mid point of B-C


And the area of the region is $\frac{1}{3}$ ABC.
$$\therefore Probability = \frac{1}{3}$$
