How do i prove that the three medians of a triangle are concurrent at a point called centroid?

beatricalwu
2022-07-27
Answered

How do i prove that the three medians of a triangle are concurrent at a point called centroid?

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juicilysv

Answered 2022-07-28
Author has **17** answers

Consider triangle ABC (below) with medians AD and BE.

Construct two points H and G as the midpoints of segments DC and BD, respectively.

It is now necessary to construct appropriate parallel lines so that we may apply the mentioned corollary. Construct four parallel lines 1, l2, l3, and l4 each parallel to line l, the line containing median AD, through pointsB,G,H, and C, respectivey. The transversal line BC has congruent segments, BG,GD,DG,HC.

Since E is the midpoint of AC, l3 intersects AC at the point E by the above corollary, using AC as the transversal.

In addition, using BE as a transversal, we have that , BF = FR = RE.

Thus the medians AD and BE intersect at R (the centroid!), a point that is two-thirds of the way from B to E, so BR = (2/3)BE.

Repeating the process using the other two pairs of medians gives the desired concurrent point R, which is the centroid of triangle.

Construct two points H and G as the midpoints of segments DC and BD, respectively.

It is now necessary to construct appropriate parallel lines so that we may apply the mentioned corollary. Construct four parallel lines 1, l2, l3, and l4 each parallel to line l, the line containing median AD, through pointsB,G,H, and C, respectivey. The transversal line BC has congruent segments, BG,GD,DG,HC.

Since E is the midpoint of AC, l3 intersects AC at the point E by the above corollary, using AC as the transversal.

In addition, using BE as a transversal, we have that , BF = FR = RE.

Thus the medians AD and BE intersect at R (the centroid!), a point that is two-thirds of the way from B to E, so BR = (2/3)BE.

Repeating the process using the other two pairs of medians gives the desired concurrent point R, which is the centroid of triangle.

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Do you see an elegant proof?

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for any $n>1$. The assertion we would like to prove is that the number of primes between $n$ and $2n$ tends to $\mathrm{\infty}$, if $n\to \mathrm{\infty}$, that is,

$\underset{n\to \mathrm{\infty}}{lim}\pi (2n)-\pi (n)=\mathrm{\infty}.$

Do you see an elegant proof?

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${I}_{a}^{2}=\frac{bc}{(b+c{)}^{2}}[(b+c{)}^{2}-{a}^{2}]$

If $I$ is the incenter, I wonder if there exist similar formula for the part $A{I}^{2}$.

${I}_{a}^{2}=\frac{bc}{(b+c{)}^{2}}[(b+c{)}^{2}-{a}^{2}]$

If $I$ is the incenter, I wonder if there exist similar formula for the part $A{I}^{2}$.