Point inside a triangle is always closer to the side of a triangle rather than to vertices of the triangle

Question

Given a triangle and a point inside the triangle, is it always the case that the point is closer (or equal distance to) one of the sides of the triangle rather than to any of the vertices of the triangle?

i.e. Given triangle $ABC$ and point $P$ inside $\triangle ABC$, prove:

$$\min\limits_{V \in {A, B, C}} d(A,V) \geq \min\limits_{E \in {AB, BC, CA}} d(A, E)$$

It may be important to note that the side of the triangle is valid to measure the distance to, if the line connecting P and the side is also inside the triangle.

Answer 1

Well, drop a perpendicular to each side. It should be obvious that the shortest of these is the shortest distance to any point on that side (including the vertices).

If the triangle is obtuse (thanks for pointing out this case!), then (at most) one perpendicular will be outside the triangle. We can eliminate this side from consideration because the distance from $P$ to the other side containing the shared vertex will be shorter than the distance to the vertex.

Answer 2

For some point $P$ in the triangle, the line connecting that point to the closest point on any side will be perpendicular to that side. If we let this closest point on a side be $D$, and let $A$ be the closest vertex to $P$, then APD will be a right-triangle with AP as its hypotenuse.

The length of AD is the shortest distance to a side, the length of AP is the shortest distance to a vertex, and we know that AD must be shorter than AP because AP is the hypotenuse, therefore it is true that a point can always be found on a side which is closer to $P$ than any vertex.