Talk:Median (geometry)

comments

Hi, I have a question about this page, does a median cut the angle of the vertex in half? Thanks, Aron G, Eh? Aron G 03:08, 2 December 2005 (UTC)[reply]

No - except when the triangle is isoceles. This is why the incenter is usually not the same as the centroid. --Henrygb 23:49, 2 December 2005 (UTC)[reply]

hey, does the median cut the distance between the two midpoint is half ? thanks Sarah

An unclear question, but if you are asking whether the point halfway from the vertex to the midpoint of the opposite side is exactly halfway between the other two side midpoints then yes. --Henrygb 03:03, 24 March 2006 (UTC)[reply]

Proving the Median creates two equal area triangles

How can you prove that the median divides a triangle into two equal areas ? —The preceding unsigned comment was added by 80.11.205.205 (talk) 18:32, 23 February 2007 (UTC).[reply]

Use the fact that the area of a triangle is half base times perpendicular height. QED --Henrygb 20:27, 23 February 2007 (UTC)[reply]

That is untrue, only an isosceles triangle or equilateral triangle can be cut into two equal areas, and only if the median is taken from the odd angle in the isosceles triangle (or any in the equilateral). Also, the proof given in the page is only for isosceles and equilateral triangles as well, assuming BAC represents the odd angle. This is because [ABE] does not equal [ACE], unless AB=AC, which it does not unless they are the similar sides in an isosceles or equilateral triangle. —Preceding unsigned comment added by 71.231.144.99 (talk) 01:33, 30 March 2011 (UTC)[reply]

The above proof by Henrygb is correct. When a median divides a triangle into two smaller triangles, both of those triangles have the same length of their base (namely half the length of the side of the original triangle), and both the triangles have the same height. So they both have the same area. For further information, see Bisection#Area_bisectors_and_area-perimeter_bisectors_of_a_triangle. Duoduoduo (talk) 14:28, 30 March 2011 (UTC)[reply]

Recent addition

The table of formulas that was added today contains useful information, but there are a number of problems with it.

1. To fit with the flow of the article, it should be merged with the pre-existing section "Formula for length", which it overlaps with.

2. The notation conflicts with the notation elsewhere in the article. The article use lower case "m_a" etc. to refer to medians, while the new section uses upper case "M_a" etc. And the new section uses lower case "m_a" etc. to refer to side half-lengths instead of to medians.

3. The caption uses the Spanish word "y" several times where it should use the English word "and".

4. The source is a Spanish-language book; while this helps to establish the accuracy of the information, it would be preferable to have an English-language source in an English Wikipedia article.

Will the editor who made the addition please merge it with the earlier section, change the notation in the table and accompanying graph, and replace "y" with "and" (and if possible find an English-language source to insert? Thanks. Duoduoduo (talk) 23:47, 13 April 2011 (UTC)[reply]

I'll go ahead and merge it with the pre-existing section, and make the notation conform. I'll delete the new graph, because it is redundant with respect to the earlier graph in the article (in addition to having the wrong notation). Duoduoduo (talk) 14:42, 14 April 2011 (UTC)[reply]

Hello i am de editor of de table "user:Gusbelluwiki" or User talk:Gusbelluwiki, dear User:Duoduoduo the four points that you expose are correct, please give me two days to fix them, I will do after the merger, as you request. Greets him cordially--Gusbelluwiki (talk) 05:08, 15 April 2011 (UTC)[reply]

User:Duoduoduo, please forget my last comment. You did an excellent job in the article "Median (geometry)".--Gusbelluwiki (talk) 09:01, 15 April 2011 (UTC)[reply]

Proof that all 3 medians intersect in a point

I was looking for a proof that all 3 medians of a triangle (i.e. lines drawn from the vertices to the midpoints of the opposite sides of the triangle) meet in a single point. I assumed this was proved somewhere in Euclid's Elements, but on skimming through his theorems on triangles I don't see it. The opening of the article here states that the 3 medians intersect in the centroid, which is a single point, but I don't see a proof that they do. Of course, any 2 medians (being straight non-parallel lines) will intersect somewhere, but it doesn't follow that all 3 will intersect in the same point. I found another Wiki article on Ceva's Theorem, which proves a more general theorem from which the intersection of the 3 medians would follow easily as a special case. But that is a rather roundabout way of proving it, and Ceva's Theorem is historically recent (i.e. not known to the Greeks), and I hope there is a simpler and more direct proof of the 3-median intersection. I just haven't found one myself! If anyone can refer me to one I would be grateful, and it would be useful to insert a proof (or reference to one) in the article.109.149.185.249 (talk) 10:16, 3 September 2018 (UTC) [Added to my previous comment: Since writing this I have found several proofs online, but most of them seem more complex than the one via Ceva's Theorem, even if that is spelled out in full. However, I did find this one which seems quite direct and neat, provided one knows the properties of similar triangles: https://mathbitsnotebook.com/Geometry/Constructions/CCCentroid.html ] 109.149.185.249 (talk) 13:07, 3 September 2018 (UTC)[reply]

nice proof.--Kmhkmh (talk) 15:06, 3 September 2018 (UTC)[reply]