Here we find out an important relationship about the triangles on the *legs* of a trapezoid.

**What’s going on:** Given a trapezoid with diagonals, prove that the area of each triangle on a leg is the geometric mean* of the areas of the triangles on the bases.

**Student handout: **Triangles-in-a-Trapezoid-3

**GeoGebra file:** here

## Teacher Notes

In the previous problem we showed the two triangles on the legs have the same area. To prove this geometric mean result we use the similar triangles on the bases. For the bases, let DC = a and AB = b. For the triangles on the bases, the ratio of the lengths* a/b (or a:b). The ratio of all lengths is a/b, including the ratios of the heights of the triangles. If the heights of the triangles are h_{1} and h_{2}** **then h_{1}/h_{2} = a/b. Also note that h_{1} + h_{2} is the height of the trapezoid (and the big triangles).

**Important/Useful Takeaways for students:**

- Similar triangles give us proportions. These are useful and easy to use. One can solve for a variable and plug into another equation.
- The geometric mean of two quantities a and b is the square root of the product.
- Geometric mean of a and b =

*The Geometric Mean is a useful mathematical concept. More information here and here. The Geometric Mean is less than or equal to the Arithmetic Mean.

In the next problem we will see another *mean, *the *harmonic mean.*