Geometric mean theorem

${\displaystyle \triangle ABC\sim \triangle ADC\sim \triangle DBC}$

Proof of theorem:

The triangles △ADC , △ BCD are similar, since:

• consider triangles △ABC, △ACD ; here we have
  $${\displaystyle \angle ACB=\angle ADC=90^{\circ },\quad \angle BAC=\angle CAD;}$$

  
  therefore by the AA postulate
  $${\displaystyle \triangle ABC\sim \triangle ACD.}$$

  
• further, consider triangles △ABC, △BCD ; here we have
  $${\displaystyle \angle ACB=\angle BDC=90^{\circ },\quad \angle ABC=\angle CBD;}$$

  
  therefore by the AA postulate
  $${\displaystyle \triangle ABC\sim \triangle BCD.}$$

  

Therefore, both triangles △ACD, △BCD are similar to △ABC and themselves, i.e.

$${\displaystyle \triangle ACD\sim \triangle ABC\sim \triangle BCD.}$$

Because of the similarity we get the following equality of ratios and its algebraic rearrangement yields the theorem:[1]

${\displaystyle {\frac {h}{p}}={\frac {q}{h}}\,\Leftrightarrow \,h^{2}=pq\,\Leftrightarrow \,h={\sqrt {pq}}\qquad (h,p,q>0)}$

Proof of converse:

For the converse we have a triangle △ABC in which ${\displaystyle h^{2}=pq}$ holds and need to show that the angle at C is a right angle. Now because of ${\displaystyle h^{2}=pq}$ we also have ${\displaystyle {\tfrac {h}{p}}={\tfrac {q}{h}}.}$ Together with ${\displaystyle \angle ADC=\angle CDB}$ the triangles △ADC, △BDC have an angle of equal size and have corresponding pairs of legs with the same ratio. This means the triangles are similar, which yields:

${\displaystyle {\begin{aligned}\angle ACB&=\angle ACD+\angle DCB\\&=\angle ACD+(90^{\circ }-\angle DBC)\\&=\angle ACD+(90^{\circ }-\angle ACD)\\&=90^{\circ }\end{aligned}}}$

Proof with the Pythagorean theorem

In the setting of the geometric mean theorem there are three right triangles △ABC, △ADC and △DBC in which the Pythagorean theorem yields:

${\displaystyle {\begin{aligned}h^{2}&=a^{2}-q^{2}\\h^{2}&=b^{2}-p^{2}\\c^{2}&=a^{2}+b^{2}\end{aligned}}}$

Adding the first 2 two equations and then using the third then leads to:

${\displaystyle {\begin{aligned}2h^{2}&=a^{2}+b^{2}-p^{2}-q^{2}\\&=c^{2}-p^{2}-q^{2}\\&=(p+q)^{2}-p^{2}-q^{2}\\&=2pq\\\therefore \ h^{2}&=pq.\end{aligned}}}$

which finally yields the formula of the geometric mean theorem.[4]

Dissecting the right triangle along its altitude h yields two similar triangles, which can be augmented and arranged in two alternative ways into a larger right triangle with perpendicular sides of lengths p + h and q + h. One such arrangement requires a square of area h² to complete it, the other a rectangle of area pq. Since both arrangements yield the same triangle, the areas of the square and the rectangle must be identical.

The square of the altitude can be transformed into an rectangle of equal area with sides p and q with the help of three shear mappings (shear mappings preserve the area):

Shear mappings with their associated fixed lines (dotted), starting with the original square as preimage each parallelogram displays the image of a shear mapping of the figure left of it