Definition D1
The segment joining a vertex of a triangle to the midpoint of the opposite side is called a median.
Proof.
(1)The midpoint of each side exists uniquely, so a triangle has exactly three medians.A6
(2)Let L, M, N be the midpoints of BC, CA, AB respectively, and write the three medians as AL, BM, CN.(1), convention
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Lemma L1
The segment joining the midpoints of two sides of a triangle is parallel to the remaining side and half its length. That is, if N, M are the midpoints of sides AB, AC, then NM ∥ BC and NM = ½BC
Proof.
(1)M, N exist uniquely and AN = ½AB, AM = ½ACA6, D1
(2)AB/AN = 2 = AC/AM, and ∠A is the same angle of △ABC and △ANM.(1), A12, A14, convention
(3)Two side ratios are equal and the included angle is the same, so △ABC and △ANM are similar under A↔A, B↔N, C↔M.(2), A3
(4)By similarity, ∠ABC = ∠ANM and BC/NM = 2(3), (2), A4
(5)The corresponding angles ∠ABC and ∠ANM are equal, so NM ∥ BC(4), A5
(6)BC/NM = 2 and NM > 0, so NM = ½BC(4), A12, A14, E
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Lemma L2
The two medians AL and BM meet at a point G between A and L and between B and M.
Proof.
(1)L is an interior point of side BC, and M of side CA.A6, D1
(2)M is an interior point of side CA, hence interior to ∠ABC, and L is an interior point of side BC, hence interior to ∠BAC.(1), A8
(3)L lies on ray BC, so ∠ABL = ∠ABC. Ray BM goes into the interior of ∠ABL in △ABL, so it meets segment AL at a point G.(1), (2), A7
(4)In the same way, ray AL meets segment BM at a point G′ in △ABM.(1), (2), A7
(5)L is not on line AB, so lines AL and BM are distinct, and two lines meet in at most one point.(1), A9, convention
(6)G and G′ both lie on the two lines, so G = G′, and this point lies in the interior of segment AL and of segment BM.(3), (4), (5)
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Lemma L3
For the point G of L2, AG = 2·GL and BG = 2·GM
Proof.
(1)M and L are the midpoints of sides CA, CB, so applying L1 with vertex C in place of A gives ML ∥ AB and ML = ½ABD1, L1, convention
(2)G is the interior intersection of segments AL and BM, so line BM cuts the two parallel lines ML and AB.L2, (1)
(3)A and L lie on opposite sides of line BM, so the alternate angles are equal. ∠GML = ∠GBA(2), L2, A1
(4)Line AL also cuts the two parallels, so for the same reason ∠GLM = ∠GAB(2), L2, A1
(5)Two angles are respectively equal, so △GML and △GBA are similar under G↔G, M↔B, L↔A.(3), (4), A2
(6)By similarity, GM/GB = GL/GA = ML/AB = ½(5), (1), A4
(7)Lengths are greater than 0, so AG = 2·GL, BG = 2·GM(6), A12, A14, E
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Theorem T
The three medians of a triangle meet at one point, and that point divides each median in the ratio 2:1 from the vertex.
Proof.
(1)Medians AL and BM meet at a point G and AG = 2·GLL2, L3
(2)Applying the same argument to medians AL and CN, the two medians meet at a point G″ and AG″ = 2·G″LL2, L3, convention
(3)G and G″ both lie in the interior of segment AL, so AG + GL = AL, AG″ + G″L = AL(1), (2), L2, A10
(4)2·GL + GL = 3·GL = AL, so GL = AL/3 and AG = 2·AL/3. The same holds for G″, so AG = AG″(1), (2), (3), A13, A14, E
(5)On ray AL there is only one point at distance AG from A, so G = G″(4), A11
(6)Hence all three medians pass through G. This point is called the centroid.(1), (2), (5), convention
(7)Applying L3 to each median gives AG = 2·GL, BG = 2·GM, CG = 2·GN(6), L3
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