Area such that $d(P, AB) \leq \min \{d(P,BC), \ d(P,AC)\}$ 
$d(P,L)$ means distance of point $P$ from line $L$.
There are three points $A(4,4)$, $B(8,4)$ and $C(4,6)$.
We need to find the area of the region satisfying $d(P,AB) \leq \min \{d(P,BC), \ d(P,AC)\}$

I drew the triangle $\triangle ABC$ and tried some geometry, but wasn't able to get it.
Solutions and hints will be appreciated.
 A: The region where $d(P,AB) \leq d(P,BC)$ is defined by the angle bisectors of the angles formed by lines $AB$ and $BC$, and so on.
In the actual region that you state, the region $R$ is given by the areas shaded yellow or red in the diagram below. The areas on the left and on the right are infinitely large, so the answer to your question is $\infty$.

If you limit the region further to the interior of the triangle, you see that $R$ is then the triangle defined by $A$, $B$, and the incenter of triangle $ABC$. (I.e. $\triangle ABI$.) The coordinates of the incenter $I$ can be found to be $(7 - \sqrt 5, 7 - \sqrt 5)$.
The area of $\triangle ABI$ is then "one-half base times height" or 
$$\frac12 \cdot 4 \cdot ((7 - \sqrt 5)-4) = 6 - 2 \sqrt 5  \approx  1.52786$$
A: HINT:
To solve the problem step by step:


*

*First, try to define the region $R_1$ where $d(P,\overline{AB})\leq d(P,\overline{AC})$

*Second, try to define the region $R_2$ where $d(P,\overline{AB})\leq d(P,\overline{BC})$

*Intersect those regions: $R=R_1\cap R_2$

*Profit!


In 1., note that the point $P$ that satisfies the condition region MUST be nearer to $\overline{AB}$ than to $\overline{AC}$. It's kind of intuitive that the region $R_1$ will be defined by the bisecting lines between the lines $\overline{AB}$ and $\overline{AC}$.
Try it!
Using the same argument of 1., you can say $R_2$ is defined by the bissecting lines between $\overline{AB}$ and $\overline{BC}$.
Then, overlap both $R_1$ and $R_2$ to get $R$.
Good luck! 
