VIDEO

The mean proportion is any value that can be expressed just the way that 'x' is in the proportion on the above on the left .

In the proportion above on the left 'x', is the geometric mean, we could solve for x by cross multiplying and going from there (more on that later)

In the proportion above on the left , '4', is the geometric mean

So what does this have to do with right similar triangles?
It turns out the when you drop an altitude (h in the picture below) from the the right angle of a right triangle , the length of the altitude becomes a geometric mean.
This occurs because you end up with similar triangles which have proportional sides and the altitude is the long leg of 1 triangle and the short leg of the other similar triangle .

Diagram 1

Interactive Demonstration
To better understand how the altitude of a right triangle acts as a mean proportion in similar triangles , look at the triangle below with sides a, b and c and altitude H.

SAAS
HLLH (Case 1)
HLLH (Case 2)

$
\frac{\class{side1}{side1}}{\class{altitude}{altitude}} = \frac{\class{altitude}{altitude}}{\class{side2}{side2}}
\\
\\
\frac{\class{side1}{BD}}{\class{altitude}{AD}} = \frac{\class{altitude}{AD}}{\class{side2}{CD}}
\\
\\
\frac{\class{side1 side1-v}{6.19}}{\class{altitude altitude-v}{6.19}} = \frac{\class{altitude altitude-v}{6.19}}{\class{side2 side2-v}{6.19}}
$

$
\frac{\class{hyp}{hyp}}{\class{leg1}{leg1}} = \frac{\class{leg1}{leg1}}{\class{side1}{side1}}
\\
\frac{\class{hyp}{BC}}{\class{leg1}{AB}} = \frac{\class{leg1}{AB}}{\class{side1}{BD}}
\\
\frac{\class{hyp hyp-v}{12.37}}{\class{leg1 leg1-v}{8.75}} = \frac{\class{leg1 leg1-v}{8.75}}{\class{side1 side1-v}{6.19}}
$

$
\frac{\class{hyp}{hyp}}{\class{leg2}{leg2}} = \frac{\class{leg2}{leg2}}{\class{side2}{side2}}
\\
\frac{\class{hyp}{BC}}{\class{leg2}{AC}} = \frac{\class{leg2}{AC}}{\class{side2}{CD}}
\\
\frac{\class{hyp hyp-v}{12.37}}{\class{leg2 leg2-v}{8.75}} = \frac{\class{leg2 leg2-v}{8.75}}{\class{side2 side2-v}{6.19}}
$

Separate Triangles
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Examples: 2 Types of Problems
Students usually have to solve 2 different core types of problems involving the geometric mean.

Problem Type 1
Problem Type 2. Hypotenuse, Leg and Side