The midsegment of a triangle and of a trapezoid

Geometry, grade 10 (4 units) Lesson 6 of 20Learning

The midsegment of a triangle and of a trapezoid

A triangle’s midsegment

A midsegment of a triangle is a segment joining the midpoints of two sides.

Theorem: a midsegment of a triangle is parallel to the third side and equal to half of it.

\displaystyle DE

\displaystyle =

\displaystyle \frac{1}{2}BC

Proving the theorem

Given: $\displaystyle D$ is the midpoint of the side $\displaystyle AB$ , and $\displaystyle E$ the midpoint of the side $\displaystyle AC$ .
Prove: $\displaystyle {DE}={\frac{1}{2}BC}$ and $\displaystyle DE\parallel BC$ .

The proof below shows that the triangle $\displaystyle ADE$ is similar to the triangle $\displaystyle ABC$ in the ratio $\displaystyle \frac{1}{2}$ (SAS), and draws the length and the parallel from the similarity.

	Claim	Reason
1	$\displaystyle {AD}={\frac{1}{2}AB}$	Given ( $\displaystyle D$ is the midpoint of $\displaystyle AB$ )
2	$\displaystyle {AE}={\frac{1}{2}AC}$	Given ( $\displaystyle E$ is the midpoint of $\displaystyle AC$ )
3	$\displaystyle {\angle DAE}={\angle BAC}$	A common angle
4	$\displaystyle \triangle ADE\sim\triangle ABC$	(1), (2), (3); the SAS similarity theorem, in the ratio $\displaystyle \frac{1}{2}$
5	$\displaystyle {DE}={\frac{1}{2}BC}$	(4); corresponding sides of similar triangles
6	$\displaystyle {\angle ADE}={\angle ABC}$	(4); corresponding angles of similar triangles
7	$\displaystyle DE\parallel BC$	(6); equal corresponding angles – the lines are parallel

The converse theorems

Two theorems recognise a midsegment:

a line through the midpoint of one side of a triangle, parallel to a second side, bisects the third side;
a segment whose ends are on two sides of a triangle, parallel to the third side and equal to half of it, is a midsegment.

Both follow from a segment parallel to a side: it cuts off a similar triangle, and the similarity ratio fixes where its ends are.

Exercise 1

In the triangle $\displaystyle ABC$ , the point $\displaystyle D$ is the midpoint of the side $\displaystyle AB$ .
The point $\displaystyle E$ is on the side $\displaystyle AC$ , and $\displaystyle DE\parallel BC$ .
Prove: $\displaystyle {AE}={EC}$ .
(A hint: $\displaystyle \triangle ADE\sim\triangle ABC$ .)

Exercise 2

The sides of a triangle are $\displaystyle 8$ , $\displaystyle 10$ and $\displaystyle 12$ long.
Its three midsegments make a small triangle.
Find the perimeter of the small triangle.

/ – 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

A trapezoid’s midsegment

The midsegment of a trapezoid joins the midpoints of its two legs.

Theorem: a trapezoid’s midsegment is parallel to the bases and equal to half their sum.

\displaystyle EF

\displaystyle =

\displaystyle \frac{a+b}{2}

Given: $\displaystyle ABCD$ is a trapezoid, $\displaystyle BC\parallel AD$ ; $\displaystyle E$ is the midpoint of the leg $\displaystyle AB$ , and $\displaystyle F$ the midpoint of the leg $\displaystyle CD$ .
Prove: $\displaystyle {EF}={\frac{1}{2}(AD+BC)}$ .

The proof below takes $\displaystyle G$ , the midpoint of the diagonal $\displaystyle AC$ , and finds two midsegments: $\displaystyle EG$ in the triangle $\displaystyle ABC$ and $\displaystyle GF$ in the triangle $\displaystyle ACD$ .

	Claim	Reason
1	$\displaystyle E$ is the midpoint of $\displaystyle AB$	Given
2	$\displaystyle G$ is the midpoint of $\displaystyle AC$	By construction
3	$\displaystyle {EG}={\frac{1}{2}BC}$	(1), (2); a midsegment of the triangle $\displaystyle ABC$
4	$\displaystyle EG\parallel BC$	(1), (2); a midsegment of the triangle $\displaystyle ABC$
5	$\displaystyle F$ is the midpoint of $\displaystyle CD$	Given
6	$\displaystyle {GF}={\frac{1}{2}AD}$	(2), (5); a midsegment of the triangle $\displaystyle ACD$
7	$\displaystyle GF\parallel AD$	(2), (5); a midsegment of the triangle $\displaystyle ACD$
8	$\displaystyle {EF}={EG+GF}$	(4), (7), $\displaystyle BC\parallel AD$ ; through $\displaystyle G$ there is only one line parallel to $\displaystyle AD$ , so $\displaystyle E$ , $\displaystyle G$ , $\displaystyle F$ are on one line
9	$\displaystyle {EF}={\frac{1}{2}(AD+BC)}$	(3), (6), (8)

Exercise 3

Calculate.

The bases of a trapezoid are $\displaystyle 7$ and $\displaystyle 13$ long.
How long is its midsegment?

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A trapezoid’s midsegment is $\displaystyle 9$ long, and one of its bases $\displaystyle 5$ .
How long is the other base?

Exercise 4

$\displaystyle ABCD$ is any quadrilateral.
$\displaystyle E$ , $\displaystyle F$ , $\displaystyle G$ and $\displaystyle H$ are the midpoints of the sides $\displaystyle AB$ , $\displaystyle BC$ , $\displaystyle CD$ and $\displaystyle DA$ .
Prove: $\displaystyle EFGH$ is a parallelogram.
(A hint: draw the diagonal $\displaystyle AC$ .)

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 $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$