Thales' Theorem
Yaritza wants to know how tall the palm tree in front of her school is, and she is not planning to climb it. She sticks a $1.5$ m stick into the ground and measures the shadows. With a little geometry and a proportion, the answer comes out. That tool is Thales' theorem, and in this section you will see what it says, why it is true, and how to use it.
Thales' theorem
In the figure, the three horizontal lines are parallel and the two transversals cut them:
Here $\dfrac{6}{4.5} = \dfrac{8}{x}$. Each segment on one transversal is compared with its corresponding one, the one between the same parallels.
Find the value of the unknown segment $x$ in the following figure, given that the horizontal lines are parallel to each other.
Find the value of the unknown segment $x$ in the following figure, given that the horizontal lines are parallel to each other.
Two intersecting lines are cut by two parallel lines. On the first line, the segments determined by the parallels measure $12$ cm and $15$ cm. On the second line, the…
Two intersecting lines are cut by two parallel lines. On the first line, the segments determined by the parallels measure $12$ cm and $15$ cm. On the second line, the segment corresponding to the $12$ cm segment measures $8$ cm. Calculate the length of the segment corresponding to the $15$ cm segment.
Find the value of $x$ using Thales' Theorem in the following diagram, where the three vertical lines are parallel. Express the answer as an integer or a simplified…
Find the value of $x$ using Thales' Theorem in the following diagram, where the three vertical lines are parallel. Express the answer as an integer or a simplified fraction.
Thales in the triangle and its proof
When a line parallel to one side cuts the other two sides of a triangle, we have the same theorem with the transversals joined at a vertex. The triangles are said to be in Thales position.
Why is it true? Proof in the triangle
Compare areas. Triangles $AMN$ and $MNB$ have the same height (the distance from $N$ to line $AB$), so their areas are in the ratio of their bases:
$$\dfrac{\text{area}(AMN)}{\text{area}(MNB)} = \dfrac{AM}{MB}$$Likewise, $AMN$ and $MNC$ have the same height (the distance from $M$ to line $AC$):
$$\dfrac{\text{area}(AMN)}{\text{area}(MNC)} = \dfrac{AN}{NC}$$But $MNB$ and $MNC$ share the base $\overline{MN}$ and, since $MN \parallel BC$, they have the same height. So they have the same area, and the two ratios above are equal: $\dfrac{AM}{MB} = \dfrac{AN}{NC}$. ∎
Also, $AMN$ and $ABC$ share angle $A$ and have equal corresponding angles, so they are similar. The equalities involving $MN$ and $BC$ follow from that.
Proof of the general theorem (several parallels)
Let $r_1 \parallel r_2 \parallel r_3$ and two transversals $s$ and $s'$. Transversal $s$ cuts them at $A$, $B$, $C$ and transversal $s'$ at $A'$, $B'$, $C'$. We want to prove that $\dfrac{AB}{BC} = \dfrac{A'B'}{B'C'}$.
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Through $A$ draw a line parallel to $s'$. It cuts $r_2$ at $P$ and $r_3$ at $Q$.
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$AA'B'P$ is a parallelogram (opposite sides parallel), so $AP = A'B'$. Likewise $PB'C'Q$ is a parallelogram, so $PQ = B'C'$.
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In triangle $ACQ$, segment $\overline{PB}$ is parallel to $\overline{QC}$ (they lie on $r_2$ and $r_3$). By Thales in the triangle: $\dfrac{AB}{BC} = \dfrac{AP}{PQ}$.
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Substituting $AP = A'B'$ and $PQ = B'C'$: $\dfrac{AB}{BC} = \dfrac{A'B'}{B'C'}$. ∎
The triangles in the figure are in Thales' position. Calculate the lengths of sides $x$ and $y$.
The triangles in the figure are in Thales' position. Calculate the lengths of sides $x$ and $y$.
In a triangle $\Delta ABC$, a line segment $MN$ is drawn parallel to side $BC$, such that $M$ lies on side $AB$ and $N$ on side $AC$. Given that $AM = 4$ cm, $MB = 2$ cm…
In a triangle $\Delta ABC$, a line segment $MN$ is drawn parallel to side $BC$, such that $M$ lies on side $AB$ and $N$ on side $AC$. Given that $AM = 4$ cm, $MB = 2$ cm and $AN = 6$ cm, calculate the length of the complete side $AC$.
You can do it on paper and send a photo with your answer.
The converse of Thales
Why?
Through $M$ draw the line parallel to $BC$ and call $N'$ the point where it cuts $AC$. By Thales, $\dfrac{AM}{MB} = \dfrac{AN'}{N'C}$. But by hypothesis also $\dfrac{AM}{MB} = \dfrac{AN}{NC}$. There is only one point on $AC$ that divides the side in that ratio, so $N' = N$. Therefore $MN$ is that parallel. ∎
This lets you check parallelism with a tape measure, with no square or protractor.
Three parallel lines are cut by two intersecting lines $r$ and $s$. On line $r$, the segments measure $3.5$ cm and $5$ cm. On line $s$, the segment corresponding to the…
Three parallel lines are cut by two intersecting lines $r$ and $s$. On line $r$, the segments measure $3.5$ cm and $5$ cm. On line $s$, the segment corresponding to the $3.5$ cm segment measures $4.2$ cm. How long is the segment corresponding to the $5$ cm segment on line $s$?
Check whether the segments determined by three lines on two intersecting lines are proportional. On the first line, they measure $15$ cm and $20$ cm. On the second line,…
Check whether the segments determined by three lines on two intersecting lines are proportional. On the first line, they measure $15$ cm and $20$ cm. On the second line, they measure $24$ cm and $32$ cm. Are the three lines parallel? Justify your answer using the converse of Thales' Theorem.
Dividing a segment into equal parts
With a straightedge and compass, Thales lets you divide any segment into $n$ equal parts without having to measure with decimals.
Explain the geometric steps required to divide a $11$ cm segment into $5$ exactly equal parts using a straightedge, compass and Thales' Theorem.
Explain the geometric steps required to divide a $11$ cm segment into $5$ exactly equal parts using a straightedge, compass and Thales' Theorem.
Thales with unknowns and X-shaped figures
Sometimes the segments in the theorem are not numbers but expressions with $x$. There are also figures where the two parallels lie on opposite sides of a crossing point.
Hourglass figure
Two transversals cross at $P$ between two parallels. Two similar triangles are formed, one pointing up and one pointing down, and their corresponding sides are proportional.
When $x^2$ appears
If the $x^2$ does not cancel when you expand, you get a quadratic equation: bring it to the form $ax^2 + bx + c = 0$, solve it, and discard the solutions that give negative or zero lengths.
Heights that grow from a vertex
If two vertical supports stand on the same slanted arm, their heights are proportional to their distance from the point where the arm meets the base (the vertex).
In a system of two intersecting lines cut by three parallel lines, the segments on the first line measure $x$ and $x+3$. On the second line, the corresponding segments…
In a system of two intersecting lines cut by three parallel lines, the segments on the first line measure $x$ and $x+3$. On the second line, the corresponding segments measure $x-2$ and $x$. Set up the proportion, solve the resulting equation, and find the values of all four segments.
In the structure of a tower crane, the jib forms a triangle with the tension cables. From a node $A$ on the jib, a vertical tie is drawn to the lower horizontal arm,…
In the structure of a tower crane, the jib forms a triangle with the tension cables. From a node $A$ on the jib, a vertical tie is drawn to the lower horizontal arm, measuring $3$ m. From a node $B$, located $4$ m further along the jib, another vertical tie is drawn measuring $4.5$ m. How far from node $A$ is the pivot point (where the two arms of the crane meet)?
You can do it on paper and send a photo with your answer.
Two parallel lines are a distance $H$ apart. Two transversal line segments cross at a point $P$ between the parallels, forming an hourglass shape. The segments of one of…
Two parallel lines are a distance $H$ apart. Two transversal line segments cross at a point $P$ between the parallels, forming an hourglass shape. The segments of one of the transversals measure $x$ and $x+5$, while the base of the top triangle measures $12$ cm and the base of the bottom triangle measures $27$ cm. Calculate the value of $x$ and the total length of that transversal.
Shadows, pinhole cameras, and light rays
Light travels in a straight line. When a ray passes through a fixed point (a pinhole, the bulb of a lamp, the Sun), the object and its shadow or image form similar triangles.
The same reasoning works for the Sun and the Moon. If the Sun (diameter $D_S$) is at a distance $d_S$ and the Moon (diameter $D_L$) exactly covers its disk at a distance $d_L$, then $\dfrac{D_L}{d_L} = \dfrac{D_S}{d_S}$.
Back to the beginning: Yaritza's stick measures $1.5$ m and its shadow $1.2$ m, and the palm tree's shadow measures $12$ m. Then $\dfrac{1.5}{1.2} = \dfrac{h}{12}$ and $h = 15$ m. The Sun's rays are parallel, so the triangles are similar.
In a pinhole camera, light enters through a small hole and projects an inverted image of an object onto the back wall. If a tree $4$ m tall is $15$ m from the hole,…
In a pinhole camera, light enters through a small hole and projects an inverted image of an object onto the back wall. If a tree $4$ m tall is $15$ m from the hole, calculate the height of the projected image $h$ if the depth of the camera is $0.3$ m.
During a solar eclipse, the Moon passes directly between the Earth and the Sun, blocking its light. Given that the diameter of the Sun is approximately $1.4 \times 10^6$…
During a solar eclipse, the Moon passes directly between the Earth and the Sun, blocking its light. Given that the diameter of the Sun is approximately $1.4 \times 10^6$ km and it is $1.5 \times 10^8$ km from the Earth, and that the Moon is $384000$ km from the Earth, calculate the approximate diameter of the Moon using similar triangles.
A theater spotlight shines on an actor, projecting his shadow onto the backdrop. The spotlight is at ground level. The actor, who is $1.80$ m tall, walks towards the…
A theater spotlight shines on an actor, projecting his shadow onto the backdrop. The spotlight is at ground level. The actor, who is $1.80$ m tall, walks towards the backdrop. When he is $4$ m from the spotlight, his shadow measures $h_1$. If he walks $2$ m further towards the backdrop, his shadow measures $h_2$. Knowing that the total distance from the spotlight to the backdrop is $10$ m, calculate the difference in height between the two projected shadows ($h_1 - h_2$).
Measuring the inaccessible: rivers, buildings, and hills
Thales lets you measure distances and heights you cannot reach. You look for a large triangle you cannot measure and a similar small one that you can.
Width of a river
Stakes $A$, $B$, $C$ are placed on one bank, with $A$ directly across from the tree $T$ on the other bank. You look for $D$ such that $T$, $B$, $D$ are in line and $\overline{CD} \parallel \overline{TA}$. Triangles $TAB$ and $DCB$ form an hourglass at $B$, so $\dfrac{TA}{CD} = \dfrac{AB}{BC}$.
The mirror method
A flat mirror is placed on the ground. Light coming from the top of the antenna reflects off it and reaches the eye at the same angle at which it struck. This forms two similar right triangles.
Two sticks for a hill
For a summit of height $H$ that you cannot reach, you use a $2$ m stick and look from the ground, lining up your eye, the tip of the stick, and the summit. You make two observations from different points, separated by a known distance.
To measure the width of a river with parallel banks, a surveyor places three stakes $A, B$, and $C$ aligned in that order on one of the banks, positioning $A$ directly…
To measure the width of a river with parallel banks, a surveyor places three stakes $A, B$, and $C$ aligned in that order on one of the banks, positioning $A$ directly opposite a tree $T$ on the other bank (so that the segment $TA$ is perpendicular to the banks), and a fourth stake $D$ such that the points $T, B$, and $D$ are aligned. Knowing that the segment $TA$ is parallel to the segment $CD$, that $AB = 8$ m, $BC = 12$ m, and the perpendicular distance from $D$ to the river is $2$ m, calculate the width of the river using similar triangles.
A flat mirror is placed on the horizontal ground. A person whose eyes are $1.60$ m above the ground stands $1.20$ m from the center of the mirror and sees the antenna of…
A flat mirror is placed on the horizontal ground. A person whose eyes are $1.60$ m above the ground stands $1.20$ m from the center of the mirror and sees the antenna of a building reflected exactly in it. If the distance from the mirror to the base of the building is $18$ m, and both the person and the building are perpendicular to the ground, what is the total height of the antenna from the ground?
To measure the height of an inaccessible mountain without using trigonometry, a surveyor uses two vertical rods $2$ m tall. He drives the first rod into the ground and,…
To measure the height of an inaccessible mountain without using trigonometry, a surveyor uses two vertical rods $2$ m tall. He drives the first rod into the ground and, lying down, visually aligns the top of the rod with the peak of the mountain. The distance from his eye to the rod is $1.5$ m. He then moves back $150$ m in a straight line, drives in the second rod, and aligns his eye with the top of the rod and the peak once again. This time, the distance from his eye to the rod is $1.6$ m. Calculate the height of the mountain.
Ramps, trajectories, and roofs
An element that moves in a straight line and crosses vertical supports (a ramp, the path of a ball, a zinc roof truss) forms triangles in Thales position with each support.
Straight trajectories
Gable roofs
A wheelchair access ramp overcomes a vertical drop of $1.2$ m over a horizontal distance of $15$ m. If an intermediate vertical support is to be placed at a horizontal…
A wheelchair access ramp overcomes a vertical drop of $1.2$ m over a horizontal distance of $15$ m. If an intermediate vertical support is to be placed at a horizontal distance of $6$ m from the start of the ramp, what must be the height of this support?
In a tennis match, a player hits the ball at a height of $2.4$ m. The ball travels in a straight line, just brushes the top of the net (which is $0.9$ m high), and lands…
In a tennis match, a player hits the ball at a height of $2.4$ m. The ball travels in a straight line, just brushes the top of the net (which is $0.9$ m high), and lands on the opponent's court. If the horizontal distance from the player to the net is $12$ m, how far from the net will the ball bounce?
In the structure of a gabled roof, a vertical reinforcing strut is placed. The base of the roof measures $10$ m in total (from the center to the eaves is $5$ m). The…
In the structure of a gabled roof, a vertical reinforcing strut is placed. The base of the roof measures $10$ m in total (from the center to the eaves is $5$ m). The maximum height at the center is $3$ m. If the strut is placed $1.5$ m from the center towards the eaves, what must its length be?
The midsegment
Proof
Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$. Then $AM = MB$ and $AN = NC$, so $\dfrac{AM}{MB} = 1 = \dfrac{AN}{NC}$. By the converse of Thales, $\overline{MN} \parallel \overline{BC}$.
By the similarity $AMN \sim ABC$, the ratio is $\dfrac{AM}{AB} = \dfrac{1}{2}$, so $MN = \dfrac{1}{2}\,BC$. ∎
In a trapezium $ABCD$ with parallel sides (bases) measuring $AB = 25$ cm and $CD = 15$ cm, a segment $MN$ is drawn parallel to the bases, with $M$ on $AD$ and $N$ on…
In a trapezium $ABCD$ with parallel sides (bases) measuring $AB = 25$ cm and $CD = 15$ cm, a segment $MN$ is drawn parallel to the bases, with $M$ on $AD$ and $N$ on $BC$. It is known that the segment $MN$ divides the non-parallel sides in the ratio $\frac{AM}{MD} = \frac{2}{3}$. Calculate the length of segment $MN$. (Hint: Draw a line parallel to one of the non-parallel sides from a vertex).
You can do it on paper and send a photo with your answer.
Prove, using the converse of Thales' Theorem, that the line segment joining the midpoints of two sides of any triangle is parallel to the third side. If the third side…
Prove, using the converse of Thales' Theorem, that the line segment joining the midpoints of two sides of any triangle is parallel to the third side. If the third side measures $L$, what is the length of this segment in terms of $L$? Justify your answer based on similar triangles.