What is the intercept theorem (Thales's theorem)?

The intercept theorem states that if two lines intersect multiple parallel lines, the segments created on one line are proportional to the corresponding segments created on the other line.

How do you find unknown segment lengths using the intercept theorem?

You set up proportional equations relating the known and unknown segment lengths and solve using cross-multiplication.

Intercept Theorem and Problems with Solutions

The intercept theorem [1], also known as Thales's theorem, is presented along with applications to problem solving.

Intercept Theorem

In all the figures below, \(BC\) is parallel to \(DE\) and both are intercepted by \(AD\) and \(AE\) with points of intersection at \(B, C, D\) and \(E\).

intercept theorem part 1.

The intercept theorem states that the lengths of the segments are proportional as follows:

\[ \dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{AC}}{\overline{AE}} = \dfrac{\overline{BC}}{\overline{DE}} \quad \quad (I) \] \[ \dfrac{\overline{AB}}{\overline{BD}} = \dfrac{\overline{AC}}{\overline{CE}} \quad \quad (II) \]

intercept theorem part 2.

\[ \dfrac{\overline{FC}}{\overline{FB}} = \dfrac{\overline{GE}}{\overline{GD}} \quad \quad (III) \]

Problems with Solutions

Problem 1: Find segment lengths given parallel lines in a triangle

Question: In the figure below \(BC\) is parallel to \(DE\). Find \(\overline{AC}\) and \(\overline{BC}\).

intercept theorem triangle problem 1

Solution:

According to the intercept theorem (I) above, we can write:

\[ \dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{BC}}{\overline{DE}} = \dfrac{\overline{AC}}{\overline{AE}} \]

We can write two equations using the above proportions:

\[ \dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{BC}}{\overline{DE}} \quad \text{(IV)} \quad \text{and} \quad \dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{AC}}{\overline{AE}} \quad \text{(V)} \]

Substitute the known lengths of segments in equation (IV):

\[ \dfrac{6}{6+2} = \dfrac{\overline{BC}}{9} \]

Use the cross product to rewrite the above equation as:

\[ 8 \times \overline{BC} = 6 \times 9 \]

Solve for \(\overline{BC}\):

\[ \overline{BC} = \frac{54}{8} = 6.75 \]

Substitute the known lengths of segments in equation (V):

\[ \dfrac{6}{6+2} = \dfrac{\overline{AC}}{\overline{AC}+3} \]

Use the cross product to rewrite the above equation as:

\[ 6 \times (\overline{AC}+3) = 8 \times \overline{AC} \]

Multiply and simplify:

\[ 6 \overline{AC} + 18 = 8 \overline{AC} \]

Group like terms:

\[ 18 = 2 \overline{AC} \]

Solve for \(\overline{AC}\):

\[ \overline{AC} = 9 \]

Note: Triangles \(ABC\) and \(ADE\) are similar, and therefore this problem can also be solved using similar triangle theorems.

Problem 2: Find multiple segment lengths with two triangles

Question: In the figure below \(BC\) is parallel to \(DE\). Find \(\overline{DG}\), \(\overline{GE}\), \(\overline{AC}\) and \(\overline{CE}\).

intercept theorem two triangles problem 2

Solution:

Using the intercept theorem (I) above, we can write:

\[ \dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{BF}}{\overline{DG}} \]

Substitute the known lengths of segments in the above equation:

\[ \dfrac{8}{8+2} = \dfrac{6}{\overline{DG}} \]

Use the cross product to rewrite the above as:

\[ 8 \times \overline{DG} = 6 \times 10 \]

Solve for \(\overline{DG}\):

\[ \overline{DG} = \frac{60}{8} = 7.5 \]

Reuse the intercept theorem (I) above to write:

\[ \dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{BC}}{\overline{DE}} \]

Substitute the known lengths of segments in the above equation:

\[ \dfrac{8}{8+2} = \dfrac{6+4}{\overline{DE}} \]

Use the cross product to rewrite the above equation as:

\[ 8 \times \overline{DE} = 10 \times 10 \]

Solve for \(\overline{DE}\):

\[ \overline{DE} = \frac{100}{8} = 12.5 \]

Note that \(\overline{DE} = \overline{DG} + \overline{GE}\). Hence:

\[ \overline{GE} = \overline{DE} - \overline{DG} = 12.5 - 7.5 = 5 \]

Reuse the intercept theorem (I) above to write:

\[ \dfrac{\overline{AC}}{\overline{FC}} = \dfrac{\overline{AE}}{\overline{GE}} \]

Note that \(\overline{AE} = \overline{AC} + \overline{CE}\). Hence the above equation may be written as:

\[ \dfrac{\overline{AC}}{4} = \dfrac{15}{5} \]

Use the cross product to rewrite the above equation as:

\[ 5 \times \overline{AC} = 4 \times 15 \]

Solve the above to find:

\[ \overline{AC} = \frac{60}{5} = 12 \]

And find \(\overline{CE}\) as follows:

\[ \overline{CE} = \overline{AE} - \overline{AC} = 15 - 12 = 3 \]

Problem 3: Determine the altitude of a pole using a light source

Question: A technician wishes to determine the altitude of a ground-mounted long pole as follows: He uses a light source at point \(B\) mounted on a structure of height 1 meter to shine a beam of light through the tops of both the short and long poles. The height of the short pole is 5 meters and the distance between the short pole and the long pole is 5 meters. The distance between the short pole and the structure supporting the light source is 2 meters. We assume that the structure supporting the light source and the two poles are all in the same plane and perpendicular to the ground. Find the altitude \(h\) of the long pole.

intercept theorem altitude of a pole problem 3

Solution:

The light structure and the two poles are perpendicular to the ground and are therefore all parallel. Hence, we use the intercept theorem (I) to write:

\[ \dfrac{\overline{OA}}{\overline{AB}} = \dfrac{\overline{OC}}{\overline{CD}} \]

Substitute the known lengths and let \(\overline{OC} = \overline{OA} + 2\):

\[ \dfrac{\overline{OA}}{1} = \dfrac{\overline{OA}+2}{5} \]

Cross multiply the above equation and rewrite as:

\[ 5 \overline{OA} = \overline{OA} + 2 \]

Solve for \(\overline{OA}\):

\[ \overline{OA} = 0.5 \, \text{meters} \]

Use the intercept theorem (I) one more time to write:

\[ \dfrac{\overline{OA}}{\overline{AB}} = \dfrac{\overline{OF}}{h} \]

Substitute the known lengths:

\[ \dfrac{0.5}{1} = \dfrac{0.5 + 2 + 5}{h} \]

Cross multiply the above equation and rewrite as:

\[ 0.5 h = 7.5 \]

Solve for \(h\) to obtain:

\[ h = 15 \, \text{meters} \]

Problem 4: Find the area of a triangle using parallel line proportions

Question: In the figure below, the area of triangle \(OAB\) is 4 square units. \(AB\) is parallel to \(CD\). Find the area of triangle \(OCD\).

intercept theorem area of a triangle problem 4

Solution:

Draw a segment through point \(O\) that is perpendicular to and intercepts segments \(AB\) and \(CD\) at points \(L\) and \(M\) respectively as shown in the figure below.

intercept theorem area of a triangle problem 4 solution

Let \(A_1\) and \(A_2\) be the areas of triangles \(OAB\) and \(ODC\) respectively and use the formula for the area of a triangle (half the product of the altitude and base) to write:

\[ A_1 = \dfrac{1}{2} \overline{OL} \times \overline{AB} \quad \text{and} \quad A_2 = \dfrac{1}{2} \overline{OM} \times \overline{CD} \]

Use the intercept theorem (I) to write:

\[ \dfrac{\overline{OM}}{\overline{OL}} = \dfrac{\overline{OD}}{\overline{OA}} = \dfrac{7.5}{2.5} = 3 \]

and

\[ \dfrac{\overline{CD}}{\overline{AB}} = \dfrac{\overline{OD}}{\overline{OA}} = \dfrac{7.5}{2.5} = 3 \]

From the above we can write that:

\[ \dfrac{\overline{OM}}{\overline{OL}} \times \dfrac{\overline{CD}}{\overline{AB}} = 3 \times 3 = 9 \]

which may be written as:

\[ \dfrac{\overline{OM} \times \overline{CD}}{\overline{OL} \times \overline{AB}} = 9 \]

Multiply the numerator and denominator in the above by \(1/2\) and write:

\[ \dfrac{\frac{1}{2}\overline{OM} \times \overline{CD}}{\frac{1}{2}\overline{OL} \times \overline{AB}} = 9 \]

The numerator and denominator in the above expression are the areas \(A_2\) and \(A_1\) as given by the formulas above, hence we may write that:

\[ \dfrac{A_2}{A_1} = 9 \]

Hence, the area \(A_2\) of triangle \(ODC\) is given by:

\[ A_2 = 9 \times A_1 = 9 \times 4 = 36 \, \text{square units} \]

Intercept Theorem

Problems with Solutions

More References and Links to Geometry Problems