# 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 $E$.
The intersect theorem states that the lengths of the segments are proportional as follows:
$\dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{AC}}{\overline{AE}} = \dfrac{\overline{BC}}{\overline{DE}} \quad \quad (I)$
$\dfrac{\overline{AB}}{\overline{BD}} = \dfrac{\overline{AC}}{\overline{CE}} \quad \quad (II)$
$\dfrac{\overline{FC}}{\overline{FB}} = \dfrac{\overline{GE}}{\overline{GD}} \quad \quad (III)$

## Problems with Solutions

Problem 1
In the figure below $BC$ is parallel to $DE$. Find $\overline{AC}$ and $\overline{BC}$

Solution to Problem 1
Acoording 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
$\dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{BC}}{\overline{DE}}$   (IV)   and       $\dfrac{\overline{AB}}{\overline{AD}} = \dfrac{\overline{AC}}{\overline{AE}}$   (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 $\overline{BC}$
$\overline{BC} = 54/8 = 6.75$

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 that the triangles $ABC$ and $ADE$ are similar and therefore this problem can also be solved using similar triangle theorem.

Problem 2
In the figure below $BC$ is parallel to $DE$. Find $\overline{DG}$, $\overline{GE}$, $\overline{AC}$ and $\overline{CE}$.

Solution to Problem 2

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 $\overline{DG}$
$\overline{DG} = 60 / 8 = 7.5$
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 $\overline{DE}$
$\overline{DE} = 100 / 8 = 12.5$

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
$\overline{AC} = 60/5 = 12$

and find $\overline{CE}$ as follows
$\overline{CE} = \overline{AE} - \overline{AC} = 15 - 12 = 3$

Problems 3
A technician wishes to determine the altitude of a ground-mounted long pole as follows (see figure below): He uses a light source at point $B$ mounted on a structure of height 1 meters 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.

Solution to Problem 3
The light structure and the two poles are perpendicular to the ground and are therefore all parallel. Hence the use of the intercept theorem (I) to write
$\dfrac{\overline{OA}}{\overline{AB}} = \dfrac{\overline{OC}}{\overline{CD}}$

Substitute the known lengths and $\overline{OC}$ by $\overline{OA} + 2$
$\dfrac{\overline{OA}}{1} = \dfrac{\overline{OA}+2}{5}$

Cross multiply the above equation and rewrite as
$5 \overline{OA} = \overline{OA}+2$

Solve for $\overline{OA}$
$\overline{OA} = 0.5$ 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
$h = 15$ meters

Problems 4
In the figure below, the area of triangle $OAB$ is 4 square units. $AB$ is parallel to $CD$. Find the area of triangle $OCD$

Solution to Problem 4
Draw a segment through point $O$ that is perpendicular to and intercept segments $AB$ and $CD$ at points $L$ and $M$ respectively as shown in the figure below.

Let $A_1$ and $A_2$ be the areas of triangles $O A B$ and $O D C$ respectively and use the formula for the area of a triangle (half the product of the altitude and base ) to write

$A_1 = \dfrac{1}{2} \overline {OL} \times \overline {AB}$ and $A_2 = \dfrac{1}{2} \overline {OM} \times \overline {CD}$

Use of 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 $A_1$ and $A_2$ as given by the formulas above, hence we may write that
$\dfrac{A_2}{ A_1} = 9$

Hence The ares $A_2$ of triangle $O D C$ is given by:
$A_2 = 9 \times A_1 = 9 \times 4 = 36$ square units.

## More References and Links to Geometry Problems

The Four Pillars of Geometry - John Stillwell - Springer; 2005th edition (Aug. 9 2005) - ISBN-10 : 0387255303
similar-triangles-examples-and-problems-with-solutions.html
Geometry Tutorials, Problems and Interactive Applets
Congruent Triangles Examples

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