The meeting point of the medians

Geometry, grade 10 (5 units) Lesson 8 of 26Learning

The meeting point of the medians

The medians meet at one point

A median of a triangle joins a vertex to the midpoint of the side opposite it.

Theorem: the three medians of a triangle meet at one point. That point divides each median in the ratio $\displaystyle 2:1$ – the part next to the vertex is twice the other part.

\displaystyle AG

\displaystyle =

\displaystyle 2GD

The meeting point of the medians is the triangle’s centroid: a triangle cut out of cardboard balances on a pin at it.

Proving the theorem

Given: $\displaystyle E$ is the midpoint of $\displaystyle AC$ and $\displaystyle F$ the midpoint of $\displaystyle AB$ ; the medians $\displaystyle BE$ and $\displaystyle CF$ meet at $\displaystyle G$ .
To prove: $\displaystyle {BG}={2GE}$ and $\displaystyle {CG}={2GF}$ .

The proof below takes $\displaystyle M$ and $\displaystyle N$ , the midpoints of $\displaystyle BG$ and $\displaystyle CG$ , and shows by two midsegments that $\displaystyle FENM$ is a parallelogram. Its diagonals bisect each other – and that cuts $\displaystyle BE$ into three equal parts.

	Claim	Reason
1	$\displaystyle FE\parallel BC$	$\displaystyle FE$ is a midsegment of the triangle $\displaystyle ABC$
2	$\displaystyle {FE}={\frac{1}{2}BC}$	A midsegment is half the third side
3	$\displaystyle MN\parallel BC$	$\displaystyle M$ , $\displaystyle N$ are the midpoints of $\displaystyle GB$ , $\displaystyle GC$ : $\displaystyle MN$ is a midsegment of the triangle $\displaystyle GBC$
4	$\displaystyle {MN}={\frac{1}{2}BC}$	A midsegment is half the third side
5	$\displaystyle {FE}={MN}$	(2), (4)
6	$\displaystyle FENM$ is a parallelogram	(1), (3), (5); a pair of opposite sides parallel and equal
7	$\displaystyle {GM}={GE}$	(6); the diagonals $\displaystyle EM$ and $\displaystyle FN$ of a parallelogram bisect each other
8	$\displaystyle {BG}={2GE}$	(7); $\displaystyle {BM}={MG}$ – $\displaystyle M$ is the midpoint of $\displaystyle BG$ – so $\displaystyle {BG}={2MG}={2GE}$
9	$\displaystyle {CG}={2GF}$	(6); in the same way, $\displaystyle {GN}={GF}$ and $\displaystyle {CN}={NG}$

So the median $\displaystyle BE$ cuts the median $\displaystyle CF$ at the point two thirds of the way from $\displaystyle C$ . The same proof, for the medians $\displaystyle BE$ and $\displaystyle AD$ , shows that $\displaystyle AD$ cuts $\displaystyle BE$ two thirds of the way from $\displaystyle B$ – at the same point $\displaystyle G$ . So all three medians pass through $\displaystyle G$ .

Exercise 1

$\displaystyle G$ is the meeting point of the medians of the triangle $\displaystyle ABC$ ; $\displaystyle AD$ and $\displaystyle BE$ are medians.

\displaystyle {AD}={12}

. Find

\displaystyle AG

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle {BG}={10}

. Find the median

\displaystyle BE

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The angle

\displaystyle C

is right, and the hypotenuse

\displaystyle {AB}={18}

. The median

\displaystyle CF

goes to the hypotenuse. Find

\displaystyle CG

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The meeting point of the medians in coordinates

In coordinates, the meeting point of the medians is the mean of the three vertices:

\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)

Why, for the $\displaystyle x$ -coordinate: $\displaystyle D$ , the midpoint of $\displaystyle BC$ , has $\displaystyle {x_D}={\frac{x_2+x_3}{2}}$ , and $\displaystyle G$ is two thirds of the way from $\displaystyle A$ to $\displaystyle D$ :

x_G=x_1+\frac{2}{3}(x_D-x_1)=\frac{1}{3}x_1+\frac{2}{3}\cdot\frac{x_2+x_3}{2}=\frac{x_1+x_2+x_3}{3}

Exercise 2

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(1,7)$ , $\displaystyle B(-3,-1)$ and $\displaystyle C(8,0)$ .

$\displaystyle D$ is the midpoint of $\displaystyle BC$ . Find it.

\displaystyle x_D

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle y_D

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find the meeting point of the medians, $\displaystyle G$ .

\displaystyle x_G

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle y_G

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find the length of the median

\displaystyle AD

, in its simplest form.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

In the triangle $\displaystyle ABC$ , the medians $\displaystyle BE$ and $\displaystyle CF$ are equal and meet at $\displaystyle G$ .
Prove: $\displaystyle {AB}={AC}$ .
(A hint: first show $\displaystyle {BG}={CG}$ and $\displaystyle {GF}={GE}$ .)

Exercise 4

$\displaystyle ABCD$ is a parallelogram, and $\displaystyle E$ is the midpoint of $\displaystyle BC$ .
$\displaystyle AE$ meets the diagonal $\displaystyle BD$ at $\displaystyle G$ .
Prove: $\displaystyle {BD}={3BG}$ .
(A hint: draw the diagonal $\displaystyle AC$ . Which triangle has $\displaystyle AE$ and $\displaystyle BG$ on two of its medians?)

Formulas: geometry, grade 10 (5 units)

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Introduction to analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
The explicit equation of a line – $\displaystyle m$ the slope, $\displaystyle b$ where it crosses the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
The general equation of a line
$\displaystyle Ax+By+C=0$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
A canonical circle – its centre at the origin, $\displaystyle R$ its radius
$\displaystyle x^{2}+y^{2}=R^{2}$
The circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$

Midsegments and the meeting point of the medians

A triangle’s midsegment – $\displaystyle DE$ is parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$
The meeting point of the medians $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Pythagoras’ theorem and its converse: the triangle is right-angled, the right angle opposite $\displaystyle c$
$\displaystyle a^{2}+b^{2}=c^{2}$

Proportion and similarity

A property of a proportion
$\displaystyle \frac{a}{b}=\frac{c}{d}\ \Rightarrow\ \frac{a+b}{b}=\frac{c+d}{d}$
Thales’ theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{DB}$ $\displaystyle =$ $\displaystyle \frac{AE}{EC}$
The extended Thales theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$
The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
Heights, medians, perimeters, the radii of the circumscribed and of the inscribed circles
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{P_1}{P_2}=\frac{R_1}{R_2}=\frac{r_1}{r_2}=k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$

The circle

An inscribed angle is half the central angle on the same arc
$\displaystyle \angle ACB$ $\displaystyle =$ $\displaystyle \frac{1}{2}\angle AOB$
An inscribed angle on the diameter $\displaystyle AB$
$\displaystyle \angle ACB=90^{\circ}$
A quadrilateral $\displaystyle ABCD$ inscribed in a circle – opposite angles
$\displaystyle \angle A+\angle C$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The angle between the tangent $\displaystyle AT$ and the chord $\displaystyle AB$ equals an inscribed angle on $\displaystyle AB$ from its other side
$\displaystyle \angle BAT$ $\displaystyle =$ $\displaystyle \angle ACB$
Two tangents from the point $\displaystyle P$
$\displaystyle PA=PB$
A quadrilateral $\displaystyle ABCD$ circumscribing a circle
$\displaystyle AB+CD$ $\displaystyle =$ $\displaystyle BC+AD$
The chords $\displaystyle AB$ and $\displaystyle CD$ meet at the point $\displaystyle E$
$\displaystyle AE\cdot EB$ $\displaystyle =$ $\displaystyle CE\cdot ED$
Two secants from the point $\displaystyle P$
$\displaystyle PA\cdot PB$ $\displaystyle =$ $\displaystyle PC\cdot PD$
A tangent $\displaystyle PT$ and a secant $\displaystyle PAB$ from the point $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$