The intercept theorem is an important elementary geometry theorem. It is very useful for solving the geometric problems related to the ratios of various line segments.

## What is the Intercept Theorem?

The theorem states that if two or more parallel lines are intersected by two self intersecting lines, then the ratios of the line segments of the first intersecting line is equal to the ratio of the similar line segments of the second intersecting line. Bit confused? Ok, let’s explain it using the following example:

In the figure above, two parallel lines **FI** and **JM** are cut by two self intersecting lines **AE** and **ON. **Two triangles are formed as a result of the intersections of the lines. According to the intercept theorem, the ratio of the line segments created from **ON **should be equal to the ratio of the lengths of the line segments created from **AE**. Therefore,

**|BG|: |GK| = |BH|:|HL|**

## The Geometry Theorem Proof

- From the above figure, it is clear that the Δ
**BGH**and Δ**BKL**are similar. - From the rules of similar triangle, we can write:

**BK/BG = BL/BH………….eqn.1**

** **

- Now, we can write,
**BK=BG+GK**and**BL=BH+HL** - So, from the
**eqn.1**we can get:

(**BG+GK)/BG = (BH+HL)/BH**

Or, **1+ (GK/BG) = 1+ (HL/BH)**

Or,** GK/BG = HL/BH**

Or, **|BG|: |GK| = |BH|:|HL|**

## Practice Problem

**Problem: **If** **the points **B, D **and **F **are the mid-points of the lines **AC, CE **and **AE** respectively, then prove that:

Area of Δ**AFB = **Area of Δ**FED**

**Solution: **

- Since the points
**F**and**B**are the mid-points of the lines**AE**and**AC**respectively, so:

**AF/FE = AB/BC **

- Hence, the lines
**BF**and**EC**must be parallel (according to the intercept theorem).

- Similarly, the lines
**BD**and**AE**are parallel.

- So, for the quadrilateral
**BDEF**all the opposite sides are parallel to each other. Hence, the quadrilateral**BDEF**is a parallelogram with**ED as one of the diagonal.**

**Hence, Area of ΔEFD = **Area of **ΔBDF……….eqn2**

- Similarly, the quadrilateral
**ABDF**is also a parallelogram with**BF**as one of the diagonal.

Hence, Area of **ΔABF = **Area of **ΔBDF………eqn3**

- From the
**Eqn.2 &3,**it can be prove that:

Area of Δ**AFB = **Area of Δ**FED**

I hope this has helped you better understand the important concept of the theorem. Knowing this will help when implementing the theorem for different geometric problems.