geometric mean theorem geometric mean theorem
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.
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.
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 \(\mathtt{h = \sqrt{pq}}\).
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.
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