I thought about this question when I thought why median and altitude of an equilateral triangle are the same.
And it seems to me it is all because the vertex is directly above the midpoint.
Though I am not able to think of any proof for Why vertex is directly above the midpoint in equilateral triangle?
When I attempted to prove I see no other reason reason than the statement mentioned above.
I thought it would help if I prove how a perpendicular divides the angle in to half and I was able to do that but later it seemed of no help.
Is there any way to prove Why vertex is directly above the midpoint in equilateral triangle?
Best Answer
In any isosceles triangle, the median and altitude from the vertex representing the third angle are always the same. This follows by:
Consideration of the two triangles formed when an altitude is dropped from the 3rd vertex. These two triangles must be congruent to each other.
Therefore, the sides opposite the equal angles are the same length.
Consequently, this vertex must be on the perpendicular bisector of the opposite side. This implies that the altitude from this vertex must bisect the opposite side. Therefore, the altitude and median from this vertex must coincide.
Now, you have the begged question: why, in an equilateral triangle do the three altitudes (AKA three medians) intersect at the same point?
Edit
As indicated by the OP, you have the following definitions:
orthocenter
The point where all altitudes intersect.
centroid
The point where the all medians intersect.
Instead of focusing on the altitude and the median of an equilateral triangle, focus instead on the perpendicular bisector of the 3rd vertex of an isosceles triangle. Assume that the triangle is such that the length of the perpendicular bisector (AKA length of the altitude) is greater than (1/2) the length of the opposite side (which is true for an equilateral triangle).
As you travel up the perpendicular bisector, from the base towards the vertex, the distance from the given point on the perpendicular bisector to the opposite vertex is continuously decreasing, while the distance from the given point to either of the other two vertices is continuously increasing. Therefore, there must be exactly one point on this perpendicular bisector that is equidistant from all three vertices. This implies that this point is on all three perpendicular bisectors.
This implies that for any isosceles triangle where the altitude from the 3rd vertex is greater than (1/2) the length of the opposite side, all three perpendicular bisectors must coincide (i.e. intersect at a single point).
What distinguishes an equilateral from other similar isosceles triangles is that since all three sides are equal, each vertex must lie on the perpendicular bisector of the opposite side. Therefore, in an equilateral triangle, all three medians (AKA altitudes, AKA perpendicular bisectors) must coincide at a specific point.