Geometric Mean Theorem

Press on the numbers to see the steps of the proof.

1

Draw right triangle \(\mathtt{ABC}\). Draw altitude \(\mathtt{\overline{CD}}\) from the right angle to the hypotenuse \(\mathtt{\overline{AB}}\).

2

\(\mathtt{m\color{gray}{\measuredangle{ACB}} = m\color{gray}{\measuredangle{ADC}}}\) and \(\mathtt{m\color{purple}{\measuredangle{BAC}} = m\color{purple}{\measuredangle{CAD}}}\).

So, \(\color{blue}{\Delta\mathtt{ABC}}\) ~ \(\color{red}{\Delta\mathtt{ADC}}\) by AA Similarity.

So, \(\color{blue}{\Delta\mathtt{ABC}}\) ~ \(\color{red}{\Delta\mathtt{ADC}}\) by AA Similarity.

3

\(\mathtt{m\color{gray}{\measuredangle{ACB}} = m\color{gray}{\measuredangle{BDC}}}\) and \(\mathtt{m\color{green}{\measuredangle{ABC}} = m\color{green}{\measuredangle{CBD}}}\).

So, \(\color{blue}{\Delta\mathtt{ABC}}\) ~ \(\color{orange}{\Delta\mathtt{BCD}}\) by AA Similarity.

So, \(\color{blue}{\Delta\mathtt{ABC}}\) ~ \(\color{orange}{\Delta\mathtt{BCD}}\) by AA Similarity.

4

Thus, \(\color{red}{\Delta\mathtt{ADC}}\) ~ \(\color{orange}{\Delta\mathtt{BCD}}\). This means that \(\mathtt{\frac{h}{p} = \frac{q}{h}}\).

\(\mathtt{h^2 = pq}\), so \(\mathtt{h = \sqrt{pq}}\).

\(\mathtt{h^2 = pq}\), so \(\mathtt{h = \sqrt{pq}}\).

5

The altitude, \(\color{red}{\mathtt{h}}\), is the geometric mean of the segment lengths (\(\color{purple}{\mathtt{p}}\) and \(\color{blue}{\mathtt{q}}\)) it creates on the hypotenuse.

Start Over