The extended Thales theorem and its converse

Geometry, grade 10 (5 units) Lesson 14 of 26Learning

The extended Thales theorem and its converse

The extended Thales theorem

The extended Thales theorem: A line parallel to one side of a triangle cuts the other two sides, or their extensions, in proportional segments.

The new case is the extensions beyond the vertex: the lines $\displaystyle BD$ and $\displaystyle CE$ cross at $\displaystyle A$ , and $\displaystyle DE\parallel BC$ .

Given: $\displaystyle A$ is on the segments $\displaystyle BD$ and $\displaystyle CE$ ; $\displaystyle DE\parallel BC$ .
To prove: $\displaystyle {\frac{AD}{AB}}={\frac{AE}{AC}}$ .

The proof below copies the triangle $\displaystyle ADE$ onto the other side of $\displaystyle A$ : $\displaystyle F$ on $\displaystyle AB$ with $\displaystyle {AF}={AD}$ , and $\displaystyle G$ on $\displaystyle AC$ with $\displaystyle {AG}={AE}$ . Then $\displaystyle FG\parallel BC$ , and Thales’ theorem holds in the triangle $\displaystyle ABC$ .

	Claim	Reason
1	$\displaystyle {AF}={AD}$	Construction
2	$\displaystyle {AG}={AE}$	Construction
3	$\displaystyle {\angle FAG}={\angle DAE}$	Vertical angles
4	$\displaystyle \triangle AFG\cong\triangle ADE$	(1), (2), (3); SAS
5	$\displaystyle {\angle AFG}={\angle ADE}$	(4); corresponding angles of congruent triangles
6	$\displaystyle {\angle ADE}={\angle ABC}$	Alternate angles between the parallel lines $\displaystyle DE$ and $\displaystyle BC$
7	$\displaystyle {\angle AFG}={\angle ABC}$	(5), (6)
8	$\displaystyle FG\parallel BC$	(7); equal corresponding angles – the lines are parallel
9	$\displaystyle {\frac{AF}{AB}}={\frac{AG}{AC}}$	(8); Thales’ theorem in the triangle $\displaystyle ABC$
10	$\displaystyle {\frac{AD}{AB}}={\frac{AE}{AC}}$	(1), (2), (9)

Exercise 1

The segments $\displaystyle AD$ and $\displaystyle BC$ cross at $\displaystyle O$ .
$\displaystyle AB\parallel CD$ .
$\displaystyle {OA}={4}$ , $\displaystyle {OD}={6}$ and $\displaystyle {OB}={5}$ .
Find $\displaystyle OC$ .

The ratio of the parallel segment

Theorem: A segment parallel to a side of a triangle, with its ends on the other two sides, is to that side as the parts it cuts off are to their whole sides:

\displaystyle \frac{DE}{BC}=\frac{AD}{AB}=\frac{AE}{AC}

Given: $\displaystyle D$ is on $\displaystyle AB$ , $\displaystyle E$ is on $\displaystyle AC$ , and $\displaystyle DE\parallel BC$ .
To prove: $\displaystyle {\frac{DE}{BC}}={\frac{AD}{AB}}$ .

The proof below draws through $\displaystyle E$ a line parallel to $\displaystyle AB$ , to the point $\displaystyle F$ on $\displaystyle BC$ . Then $\displaystyle {DE}={BF}$ , and Thales’ theorem at the vertex $\displaystyle C$ gives the ratio of $\displaystyle BF$ to $\displaystyle BC$ .

	Claim	Reason
1	$\displaystyle DE\parallel BF$	Given ( $\displaystyle DE\parallel BC$ )
2	$\displaystyle EF\parallel DB$	Construction
3	$\displaystyle DBFE$ is a parallelogram	(1), (2); a quadrilateral whose opposite sides are parallel
4	$\displaystyle {DE}={BF}$	(3); opposite sides of a parallelogram
5	$\displaystyle {\frac{CF}{CB}}={\frac{CE}{CA}}$	(2); Thales’ theorem at the vertex $\displaystyle C$
6	$\displaystyle {\frac{BF}{BC}}={\frac{AE}{AC}}$	(5); subtracting both sides from $\displaystyle 1$
7	$\displaystyle {\frac{AE}{AC}}={\frac{AD}{AB}}$	Given $\displaystyle DE\parallel BC$ ; Thales’ theorem at the vertex $\displaystyle A$
8	$\displaystyle {\frac{DE}{BC}}={\frac{AD}{AB}}$	(4), (6), (7)

The same holds when $\displaystyle D$ and $\displaystyle E$ are on the extensions of the sides – beyond $\displaystyle B$ and $\displaystyle C$ or beyond $\displaystyle A$ .

Exercise 2

In the triangle $\displaystyle ABC$ , $\displaystyle D$ is on the side $\displaystyle AB$ and $\displaystyle E$ is on the side $\displaystyle AC$ .
$\displaystyle DE\parallel BC$ .
$\displaystyle {AD}={4}$ , $\displaystyle {DB}={2}$ and $\displaystyle {BC}={9}$ .
Find $\displaystyle DE$ .

The converse of Thales’ theorem

The converse of Thales’ theorem: If a line cuts two sides of a triangle in proportional segments, it is parallel to the third side.

In the triangle $\displaystyle ABC$ : if $\displaystyle D$ is on $\displaystyle AB$ , $\displaystyle E$ is on $\displaystyle AC$ and $\displaystyle {\frac{AD}{DB}}={\frac{AE}{EC}}$ , then $\displaystyle DE\parallel BC$ .

The proof below shows that the parallel to $\displaystyle BC$ through $\displaystyle D$ reaches $\displaystyle AC$ exactly at $\displaystyle E$ :

through $\displaystyle D$ draw the line parallel to $\displaystyle BC$ ; it meets $\displaystyle AC$ at a point $\displaystyle E'$ ;
by Thales’ theorem $\displaystyle {\frac{AD}{DB}}={\frac{AE'}{E'C}}$ , so $\displaystyle {\frac{AE'}{E'C}}={\frac{AE}{EC}}$ ;
adding $\displaystyle 1$ to both sides: $\displaystyle {\frac{AC}{E'C}}={\frac{AC}{EC}}$ , so $\displaystyle {E'C}={EC}$ ;
$\displaystyle E'$ and $\displaystyle E$ are both on the segment $\displaystyle AC$ , at the same distance from $\displaystyle C$ – they are one point, and $\displaystyle DE$ is the parallel to $\displaystyle BC$ .

The converse holds for the ratio of a part to its whole side too: if $\displaystyle {\frac{AD}{AB}}={\frac{AE}{AC}}$ , then $\displaystyle DE\parallel BC$ .

Exercise 3

In the triangle $\displaystyle ABC$ , $\displaystyle D$ is on the side $\displaystyle AB$ and $\displaystyle E$ is on the side $\displaystyle AC$ .
$\displaystyle {\frac{AD}{DB}}={\frac{AE}{EC}}$ .
Prove: $\displaystyle DE\parallel BC$ .

Exercise 4

$\displaystyle ABCD$ is a trapezoid, $\displaystyle AB\parallel DC$ .
Its diagonals cross at $\displaystyle O$ .
Prove: $\displaystyle {\frac{AO}{OC}}={\frac{BO}{OD}}$ .

Exercise 5

In the triangle $\displaystyle ABC$ , $\displaystyle D$ is on the side $\displaystyle AB$ and $\displaystyle E$ is on the side $\displaystyle AC$ .
Is $\displaystyle DE$ parallel to $\displaystyle BC$ ? Explain.

$\displaystyle {AD}={6}$ , $\displaystyle {DB}={4}$ , $\displaystyle {AE}={9}$ , $\displaystyle {EC}={6}$ .
$\displaystyle {AD}={5}$ , $\displaystyle {AB}={8}$ , $\displaystyle {AE}={6}$ , $\displaystyle {AC}={10}$ .

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 $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$