The medians and where they meet

Geometry, grade 10 (4 units) Lesson 16 of 20Learning

The medians and where they meet

Medians of a triangle

A median of a triangle is a segment from a vertex to the midpoint of the side opposite it. Every triangle has three medians, and from grade 8 you know:

the three medians meet at one point – where the medians meet, $\displaystyle G$ (the triangle’s centre of gravity);
in a right-angled triangle, the median to the hypotenuse is half the hypotenuse (grade 9).

The new theorem says where $\displaystyle G$ is on each median:

Theorem: the point where the medians meet divides each median in the ratio $\displaystyle 2:1$ – the part nearer the vertex is twice the other part.

\displaystyle AG

\displaystyle =

\displaystyle 2GD

Proving the ratio 2:1

Given: $\displaystyle BE$ and $\displaystyle CF$ are medians of the triangle $\displaystyle ABC$ , and they cross at the point $\displaystyle G$ .
Prove: $\displaystyle {BG}={2GE}$ .

The proof below draws the midsegment $\displaystyle FE$ , and shows that with the medians it makes a triangle $\displaystyle GFE$ similar to the triangle $\displaystyle GCB$ in the ratio $\displaystyle \frac{1}{2}$ .

	Claim	Reason
1	$\displaystyle {FE}={\frac{1}{2}BC}$	$\displaystyle F$ , $\displaystyle E$ are the midpoints of $\displaystyle AB$ , $\displaystyle AC$ ; a midsegment of a triangle
2	$\displaystyle FE\parallel BC$	A midsegment of a triangle
3	$\displaystyle {\angle GFE}={\angle GCB}$	(2); alternate angles between parallel lines
4	$\displaystyle {\angle GEF}={\angle GBC}$	(2); alternate angles between parallel lines
5	$\displaystyle \triangle GFE\sim\triangle GCB$	(3), (4); the AA similarity theorem
6	$\displaystyle {GE}={\frac{1}{2}GB}$	(1), (5); the similarity ratio $\displaystyle {\frac{FE}{BC}}={\frac{1}{2}}$
7	$\displaystyle {BG}={2GE}$	(6)

The same proof gives $\displaystyle {CG}={2GF}$ . The third median, from $\displaystyle A$ , crosses $\displaystyle BE$ at the point that divides it in the ratio $\displaystyle 2:1$ – and that is the same point $\displaystyle G$ . So it also follows that the three medians meet at one point.

Exercise 1

$\displaystyle AD$ and $\displaystyle BE$ are medians of the triangle $\displaystyle ABC$ , and they cross at the point $\displaystyle G$ .
Prove: $\displaystyle {AG}={2GD}$ .
(A hint: the midsegment $\displaystyle DE$ .)

Exercise 2

$\displaystyle G$ is where the medians of the triangle $\displaystyle ABC$ meet.
$\displaystyle AD$ 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 {GE}={5}

; find

\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 median and the triangle’s area

The median $\displaystyle AD$ splits the triangle $\displaystyle ABC$ into two triangles, $\displaystyle ABD$ and $\displaystyle ACD$ , of equal area:

their bases are equal: $\displaystyle {BD}={DC}$ ;
the altitude from $\displaystyle A$ is common to both.

\displaystyle S_{ABD}

\displaystyle =

\displaystyle S_{ACD}

Exercise 3

The area of the triangle $\displaystyle ABC$ is $\displaystyle 36$ .
$\displaystyle AD$ is a median, and $\displaystyle G$ is where the medians meet.

The area of the triangle

\displaystyle ABD

/ – 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 area of the triangle

\displaystyle GBD

(a hint:

\displaystyle {GD}={\frac{1}{3}AD}

)

/ – 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

Where the medians meet, in coordinates

$\displaystyle G$ is on the median $\displaystyle AD$ , two thirds of the way from $\displaystyle A$ to $\displaystyle D$ . Working out its coordinates gives 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)

Example: $\displaystyle A(0,0)$ , $\displaystyle B(6,0)$ , $\displaystyle C(3,9)$ .

$\displaystyle {G\left(\frac{0+6+3}{3},\frac{0+0+9}{3}\right)}={G(3,3)}$ ;
a check: $\displaystyle D$ , the midpoint of $\displaystyle BC$ , is $\displaystyle (4.5,4.5)$ ; from $\displaystyle A$ to $\displaystyle D$ is $\displaystyle 4.5$ to the right and $\displaystyle 4.5$ up, and two thirds of that – $\displaystyle 3$ and $\displaystyle 3$ .

Exercise 4

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(1,1)$ , $\displaystyle B(7,3)$ and $\displaystyle C(4,8)$ .

Write on the board the equation of the median from $\displaystyle A$ and the equation of the median from $\displaystyle B$ .

\displaystyle A=(1,1)

\displaystyle B=(7,3)

\displaystyle C=(4,8)

Find the point where the two medians meet.

Find $\displaystyle G$ by the formula, and compare.

\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

Formulas: geometry, grade 10 (4 units)

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Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
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
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
A line’s equation in explicit form – $\displaystyle m$ the slope, $\displaystyle b$ where it meets the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
A line’s equation in implicit form
$\displaystyle Ax+By+C=0$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The slope – $\displaystyle \alpha$ the angle between the line and the positive direction of the $\displaystyle x$ -axis
$\displaystyle m$ $\displaystyle =$ $\displaystyle \tan\alpha$

Similar triangles and midsegments

The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
The ratio of the perimeters
$\displaystyle \frac{P_1}{P_2}$ $\displaystyle =$ $\displaystyle k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$
The ratio of corresponding altitudes, medians and angle bisectors
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k$
A triangle’s midsegment – $\displaystyle DE$ parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$

Trigonometry

Sine – $\displaystyle a$ the side opposite the angle, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
Cosine – $\displaystyle b$ the side adjacent to the angle
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
Tangent – the opposite side over the adjacent side
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The tangent from the sine and the cosine
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
The identity from Pythagoras
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
Angles whose sum is $\displaystyle 90^{\circ}$
$\displaystyle \sin(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
A triangle’s area – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Exact values

$\displaystyle \sin30^{\circ}=\cos60^{\circ}=\frac{1}{2}$
$\displaystyle \sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin60^{\circ}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^{\circ}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{3}$
$\displaystyle \tan45^{\circ}$ $\displaystyle =$ $\displaystyle 1$
$\displaystyle \tan60^{\circ}$ $\displaystyle =$ $\displaystyle \sqrt{3}$

Special lines in a triangle

Where the medians meet, $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Where the medians meet, in coordinates
$\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$
The median $\displaystyle AD$ splits the triangle into two triangles of equal area
$\displaystyle S_{ABD}$ $\displaystyle =$ $\displaystyle S_{ACD}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$