If it true that the three medians of a triangle always intersect at a single point? If so, how to prove this?

Is it also the same for altitudes, angle bisectors, and perpendicular bisectors? If so, how to prove this?If it true that the three medians of a triangle always intersect at a single point? If so, how to prove this?
The proof for each of the three cases isn't the same. For example, the proof that angle bisectors meet at the same point is probably the easiest to visualize, it's simply the center of the circle inscribed inside a triangle. The proof for medians meeting at a common point is probably most easily done through vector analysis: Given vectors A and B which are 2 sides of a triange, then (1/2)(A+B) is the vector of the median, and (1/3)(A+B) is the vector of the intersection of the medians. Given that the 3rd side is (A-B), it's not hard to show that it's the same point for all medians. Finally, the proof for altitudes is probably the least obvious, but there are several well known proofs, as given in the link below:





It should be added that for a typical triangle, the three meeting points may not be the same.





Addenum: absird, what I don't see is how does it follow that the perpendicular bisectors of a triangle meet a common point? If we already know that the alitudes of a triangle meet at a common point, then the perpendicular bisectors must also. But how did you ';make the leap';, and say that the perpendicular bisectors must meet a point, from which you THEN show that the altitudes of a triangle must also?If it true that the three medians of a triangle always intersect at a single point? If so, how to prove this?
The three medians are concurrent at a point known as the triangle's centroid, which or centre of mass of the triangle. Note that this means that the centroid is always in the interior of the triangle. Two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side.


Proof: It can be pr oven that the three medians are concurrent using Ceva's theorem. Since the medians of a triangle pass through the midpoint of the sides, it's clear that ratio between the two segments they are divided into have ratios of 1. Thus, since , the three medians are concurrent. The latter point mentioned can be pr oven using the mid segment theorem.


go to yahoo search and type median of triangle.





The theorem was proved by Giovanni Ceva in his 1678 work De lineis rectis, but it was also proved much earlier by Al-Mu'taman ibn H疟d, an eleventh-century king of Saragossa.
I'll challenge you to come up with a proof showing that the perpendicular bisectors are concurrent at the circumcenter (just keep in mind that you wish to show the intersection point is equidistant from the three vertices).





I'll offer a proof of the concurrency for altitudes that draws upon the concurrency for perpendicular bisectors above. Consider some triangle ABC. Draw lines through each vertex parallel to the opposite sides of ABC. These parallel lines form a bigger triangle which we can call EFD. Let's have AC || DE, BC || DF, and AB || FE.





Let x be the altitude of ABC through vertex A. Since BC || DF, we have that x is perpendicular to DF. From parallelograms AFCB and ACBD we gather that AF = BC = AD, so AF = AD. Thus x is the perpendicular bisector of DF. Likewise the other two altitudes of ABC are perpendicular bisectors of FE and ED. Since the perpendicular bisectors of DEF are synonymous with the altitudes of ABC and since the perpendicular bisectors intersect at a single point so must the altitudes of ABC (they intersect at the circumcenter of DEF).





Addition: Thank you Scythian for pointing that out. You are right, of course, I didn't show that the perpendicular bisectors are concurrent. I intended that as an excercise for the asker. My work assumes that we already know that the perpendicular bisectors are concurrent.