In this section, you will practice with twocolumn proofs involving
the Pythagorean Theorem, triangle congruence theorems, and other tools.
A couple lengthy proofs are explored—you can print
worksheets for these proofs, and practice supplying reasons for each step yourself.
The first proof in this lesson involves an ‘averaging’ method
called the geometric mean,
which has important applications in (for example) biology and finance.
There are different types of ‘averaging’ methods in mathematics.
Any type of ‘average’ is a way to replace a collection of numbers
with a single number which, in some way, represents the collection.
Different ‘averaging’ methods are appropriate in different situations.
The first ‘average’ one usually encounters is the
arithmetic mean:
to find the arithmetic mean of
$\,N\,$ numbers, add them up and then divide by $\,N\,$.
The geometric mean, on the other hand, involves multiplying and taking roots:
Let $\,a\,$ and $\,b\,$ be positive numbers.
The geometric mean of $\,a\,$ and $\,b\,$
is the positive number
$\,\sqrt{ab}\,$.
Thus, to find the geometric mean of two positive numbers:
To find the geometric mean of  three  positive numbers:  multiply them together; take the  cube root  of the result. 
To find the geometric mean of  four  positive numbers:  multiply them together; take the  fourth root  of the result. 
To find the geometric mean of  $\,n\,$  positive numbers:  multiply them together; take the  $\,n^{\text{th}}\,$ root  of the result. 
Take a rectangle with sides $\,a\,$ and $\,b\,$, and reshape it into a square, without changing the area.
Then, the side length of the square is the geometric mean of $\,a\,$ and $\,b\,$:
$$
\begin{gather}
\text{area of rectangle is } ab\cr
\text{area of square is } (\sqrt{ab})(\sqrt{ab}) = ab
\end{gather}
$$
There are equivalent ways to view the geometric mean.
Here is the formulation that naturally appears in our first proof:
Note: Crossmultiplying
$\,\frac{a}{g} = \frac{g}{b}\,$ gives
$\,g^2 = ab\,$, from which $\,g = \pm\sqrt{ab}\,$.
So, indeed, the positive solution is the geometric mean of $\,a\,$ and $\,b\,$.
The geometric mean tends to ‘dampen’ the effect of very large numbers.
For example, the arithmetic mean of $\,10\,$ and $\,100,000\,$ is $\,\frac{10 + 100,000}{2} = 50,005\,$.
However, the geometric mean of $\,10\,$ and $\,100,000\,$ is only $\,\sqrt{(10)(100,000)} = 1000\,$.
This ‘dampening’ property makes the geometric mean particularly useful
when working with collections of numbers that have greatly varying sizes.
The reason this ‘dampening’ occurs is apparent from the next characterization;
it will only be accessible to you if you've already studied
logarithms and properties of logarithms.
The geometric mean makes an interesting appearance in right triangles.
If you drop an altitude from the rightangle vertex to the hypotenuse,
the hypotenuse is thus split into two segments.
The length of the altitude is the geometric mean between these two segments!
And, as the next proof shows, the converse is also true:
if an altitude of a triangle is the geometric mean between the two segments of the side it hits,
then the triangle must be a right triangle.
Sometimes, substitutions are a bit difficult to follow.
Coloring is included in the beginning of the next proof to help you see what is being substituted for what.
A worksheet follows the proof; you can practice supplying reasons for each step.
GIVEN: $\overline{CB}\perp \overline{AD}$ (that is, $\overline{CB}\,$ is an altitude of the triangle) $\displaystyle\frac{AB}{BC} =\frac{BC}{BD}$ (that is, $\,BC\,$ is the geometric mean between $\,AB\,$ and $\,BD\,$) PROVE: $\angle ACD\,$ is a right angle 

STRATEGY: Show that $\,(AC)^2 + (CD)^2 = (AB+BD)^2$ 

PROOF:  
STATEMENTS  REASONS 
1. $\,\overline{CB} \perp \overline{AD}$  given 
2. $\,\Delta ABC\,$ is a right triangle with hypotenuse $\,\overline{AC}\,$  definitions: right triangle, hypotenuse 
3. $\,\color{red}{(AB)^2 + (BC)^2 = (AC)^2}$  the Pythagorean Theorem 
4. $\,\Delta DBC\,$ is a right triangle with hypotenuse $\,\overline{CD}$  definitions: right triangle, hypotenuse 
5. $\,\color{blue}{(BD)^2+ (BC)^2 = (CD)^2}$  the Pythagorean Theorem 
6. $\,\color{red}{(AC)^2}$ $+$ $\color{blue}{(CD)^2}$ $=$ $\color{red}{(AB)^2 + (BC)^2}$ $+$ $\color{blue}{(BD)^2 + (BC)^2}$  substitution (steps 3 and 5) 
7. $\,(AC)^2+(CD)^2=(AB)^2+2(BC)^2+(BD)^2$  combine like terms 
8. $\displaystyle\,\frac{AB}{BC} =\frac{BC}{BD}$  given 
9. $\,(AB)(BD)= (BC)^2$ 
Multiplication Property of Equality; multiply both sides by $\,(BC)(BD)\,$ (i.e., crossmultiply) 
10. $\,2(AB)(BD)=2(BC)^2\,$ 
Multiplication Property of Equality; multiply both sides by $\,2$ 
11. $\,(AC)^2+(CD)^2=(AB)^2+2(AB)(BD)+(BD)^2$  substitution (steps 7 and 10) 
12. $\,(AB+BD)^2=(AB)^2+2(AB)(BD)+(BD)^2$  FOIL 
13. $\,(AC)^2+(CD)^2=(AB+BD)^2$  substitution (steps 11 and 12) 
14. $\,AB+BD=AD$  $\,B\,$ is between $\,A\,$ and $\,D\,$ 
15. $\,(AC)^2+(CD)^2=(AD)^2$  substitution (steps 13 and 14) 
16.
$\,\Delta ACD\,$
is a right triangle with hypotenuse $\,\overline{AD}\,$; so, $\,\angle ACD\,$ is a right angle 
the Pythagorean Theorem 
Here is a second proof, which gives you practice with some different tools.
Again, you can create a worksheet with a blank ‘Reasons’ column, so you can supply the reasons yourself.
GIVEN: $AB=AC$ $D\,$ is the midpoint of $\,\overline{AB}$ $E\,$ is the midpoint of $\,\overline{AC}$ PROVE: $DF=EF$ 

STRATEGY: Show that $\,\Delta BDF\cong \Delta CEF\,$ 

PROOF:  
STATEMENTS  REASONS 
1. $\,AB=AC\,$ $\,D\,$ is the midpoint of $\,\overline{AB}$ $\,E\,$ is the midpoint of $\,\overline{AC}$ 
given 
2.
$\,m\angle ACB= m\angle ABC\,$ 
angles opposite equal sides have equal measures 
3. $\,BD=\frac{1}{2}AB\,$ and $\,CE=\frac{1}{2}AC\,$  definition of midpoint 
4. $\,BD=CE\,$  substitution (steps 1 and 3) 
5. $\,\overline{BC}\cong \overline{CB}$  reflexive property (congruence is an equivalence relation on the set of geometric figures) 
6. $\,\Delta DBC\cong \Delta ECB$  SAS (steps 2, 4, 5) 
7. $\,m\angle BCD =m\angle CBE\,$  CPCTC 
8. $\,BF=CF\,$  sides opposite equal measure angles are equal 
9.
$\,m\angle ABE + m\angle CBE = m\angle ABC\,$ and $m\angle ACD + m\angle BCD = m\angle ACB$ 
angle addition 
10. $\,m\angle ABE + m\angle BCD = m\angle ACD + m\angle BCD$  substitution (steps 2, 7, and 9) 
11. $\,m\angle ABE = m\angle ACD$ 
addition property of equality (subtract $\,m\angle BCD\,$ from both sides) 
12.
$\,\angle DFB\cong \angle EFC$ 
vertical angles are congruent 
13.
$\,\Delta BDF\cong \Delta CEF$ 
ASA (steps 8, 11, 12) 
14.
$\,DF=EF\,$ (or, $\,\overline{DF} \cong \overline{EF}$ ) 
CPCTC 
In the exercises, you will also practice doing shorter twocolumn proofs;
you will need to supply both the statements and reasons.
In your proofs, feel free to use statements like
$\,\overline{AB}\cong \overline{CD}\,$ and
$\,AB=CD\,$
totally interchangeably,
since two segments are congruent if and only if they have the same length.
However, be careful to use the verb ‘$\,\cong \,$’ to compare geometric figures,
and the verb ‘$\,=\,$’ to compare numbers.
Similarly, feel free to use statements like
$\,\angle ABC\cong \angle DEF\,$ and $\,m\angle ABC = m\angle DEF\,$
totally interchangeably,
since two angles are congruent if and only if they have the same measure.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
