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# On the geometric mean theorem III

What’s new? Exciting news (from our point of view).

Writing for the third time on the geometric mean theorem, after an external eye scrutinized and stamped the previous two writings on the same theme, seems a joy thing.

With that said, today’s post focuses on numerical problems that concretize the geometric mean theorem in obtuse triangles.

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Consider with vertices at , , and . Let be the foot of the altitude from vertex . Verify that the altitude is the geometric mean of and .

Note that the triangle in this first example is a 30-30-120 triangle; it satisfies the geometric mean theorem twice.

From the above diagram, . Also and . Thus:

Similarly, if denotes the foot of the altitude from vertex , then:

Another unique property of the triangle.

Consider with vertices at , , and . If is the foot of the altitude from vertex , verify that .

From the above diagram, . And so:

In words: the altitude is the geometric mean of the segments and .

Consider with vertices at , , and . If is the foot of the altitude from vertex , verify that .

From the above diagram, . And so:

In words: the altitude is the geometric mean of the segments and .

Consider with vertices at , , and . If is the foot of the altitude from vertex , verify that .

From the above diagram, . And so:

In words: the altitude is the geometric mean of the segments and .

Consider with vertices at , , and . If is the foot of the altitude from vertex , verify that .

From the above diagram, . And so:

In words: the altitude is the geometric mean of the segments and .

## Takeaway

The two statements below, the geometric mean theorem, and (more than) 75 other statements, are equivalent in any non-right triangle with circumradius and side-lenths :

1. .