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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