PRACTICE WITH TWO-COLUMN PROOFS

In this section, you will practice with two-column 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 Geometric Mean

The first ‘average’ one usually encounters is the arithmetic mean:
to find the arithmetic mean of [beautiful math coming... please be patient] $\,N\,$ numbers, add them up and then divide by $\,N\,$.

The geometric mean, on the other hand, involves multiplying and taking roots:

DEFINITION geometric mean of two positive numbers

Let [beautiful math coming... please be patient] $\,a\,$ and $\,b\,$ be positive numbers.

The geometric mean of $\,a\,$ and $\,b\,$ is the positive number [beautiful math coming... please be patient] $\,\sqrt{ab}\,$.

Thus, to find the geometric mean of two positive numbers:

  • multiply the numbers together
  • take the square root of the result

Comments on the geometric mean

There are equivalent ways to view the geometric mean.
Here is the formulation that naturally appears in our first proof:

Equivalent Characterization of the geometric mean

The geometric mean of $\,a\,$ and $\,b\,$ is the positive solution $\,g\,$ of the equation [beautiful math coming... please be patient] $\displaystyle\,\frac{a}{g} = \frac{g}{b}\,$.

The number $\,g\,$ is also referred to as the geometric mean between $\,a\,$ and $\,b\,$.

Note:   Cross-multiplying [beautiful math coming... please be patient] $\,\frac{a}{g} = \frac{g}{b}\,$ 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.

Equivalent Characterization of the geometric mean

The geometric mean of positive numbers $\,a\,$ and $\,b\,$ can be computed as follows:
PROOF #1: Right Triangles and the Geometric Mean

The geometric mean makes an interesting appearance in right triangles.
If you drop an altitude from the right-angle 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:

[beautiful math coming... please be patient] $\overline{CB}\perp \overline{AD}$   (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 [beautiful math coming... please be patient] $\,(AC)^2 + (CD)^2 = (AB+BD)^2$
PROOF:
STATEMENTSREASONS
1.   [beautiful math coming... please be patient] $\,\overline{CB} \perp \overline{AD}$ given
2.   [beautiful math coming... please be patient] $\,\Delta ABC\,$ is a right triangle with hypotenuse $\,\overline{AC}\,$ definitions: right triangle, hypotenuse
3.   $\,(AB)^2 + (BC)^2 = (AC)^2$ the Pythagorean Theorem
4.   [beautiful math coming... please be patient] $\,\Delta DBC\,$ is a right triangle with hypotenuse $\,\overline{CD}$ definitions: right triangle, hypotenuse
5.   $\,(BD)^2+ (BC)^2 = (CD)^2$ the Pythagorean Theorem
6.   $\,(AC)^2$ $+$ $(CD)^2$ $=$ $(AB)^2 + (BC)^2$ $+$ $(BD)^2 + (BC)^2$ substitution (steps 3 and 5)
7.   [beautiful math coming... please be patient] $\,(AC)^2+(CD)^2=(AB)^2+2(BC)^2+(BD)^2$ combine like terms
8.   [beautiful math coming... please be patient] $\displaystyle\,\frac{AB}{BC} =\frac{BC}{BD}$ given
9.   [beautiful math coming... please be patient] $\,(AB)(BD)= (BC)^2$ Multiplication Property of Equality;
multiply both sides by $\,(BC)(BD)\,$
(i.e., cross-multiply)
10.   $\,2(AB)(BD)=2(BC)^2\,$ Multiplication Property of Equality;
multiply both sides by $\,2$
11.   [beautiful math coming... please be patient] $\,(AC)^2+(CD)^2=(AB)^2+2(AB)(BD)+(BD)^2$ substitution (steps 7 and 10)
12.   [beautiful math coming... please be patient] $\,(AB+BD)^2=(AB)^2+2(AB)(BD)+(BD)^2$ FOIL
13.   [beautiful math coming... please be patient] $\,(AC)^2+(CD)^2=(AB+BD)^2$ substitution (steps 11 and 12)
14.   $\,AB+BD=AD$ $\,B\,$ is between $\,A\,$ and $\,D\,$
15.   [beautiful math coming... please be patient] $\,(AC)^2+(CD)^2=(AD)^2$ substitution (steps 13 and 14)
16.   [beautiful math coming... please be patient] $\,\Delta ACD\,$ is a right triangle with hypotenuse $\,\overline{AD}\,$;
so, $\,\angle ACD\,$ is a right angle
the Pythagorean Theorem

PROOF #2

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:

[beautiful math coming... please be patient] $AB=AC$
$D\,$ is the midpoint of $\,\overline{AB}$
$E\,$ is the midpoint of $\,\overline{AC}$

PROVE:

$DF=EF$
STRATEGY:
Show that [beautiful math coming... please be patient] $\,\Delta BDF\cong \Delta CEF\,$
PROOF:
STATEMENTSREASONS
1.   $\,AB=AC\,$
[beautiful math coming... please be patient] $\,D\,$ is the midpoint of $\,\overline{AB}$
$\,E\,$ is the midpoint of $\,\overline{AC}$
given
2.   [beautiful math coming... please be patient] $\,m\angle ACB= m\angle ABC\,$
angles opposite equal sides have equal measures
3.   [beautiful math coming... please be patient] $\,BD=\frac{1}{2}AB\,$ and $\,CE=\frac{1}{2}AC\,$ definition of midpoint
4.   [beautiful math coming... please be patient] $\,BD=CE\,$ substitution (steps 1 and 3)
5.   [beautiful math coming... please be patient] $\,\overline{BC}\cong \overline{CB}$ reflexive property (congruence is an equivalence relation on the set of geometric figures)
6.   [beautiful math coming... please be patient] $\,\Delta DBC\cong \Delta ECB$ SAS (steps 2, 4, 5)
7.   [beautiful math coming... please be patient] $\,m\angle BCD =m\angle CBE\,$ CPCTC
8.   [beautiful math coming... please be patient] $\,BF=CF\,$ sides opposite equal measure angles are equal
9.   [beautiful math coming... please be patient] $\,m\angle ABE + m\angle CBE = m\angle ABC\,$   and
      $m\angle ACD + m\angle BCD = m\angle ACB$
angle addition
10.   [beautiful math coming... please be patient] $\,m\angle ABE + m\angle BCD = m\angle ACD + m\angle BCD$ substitution (steps 2, 7, and 9)
11.   [beautiful math coming... please be patient] $\,m\angle ABE = m\angle ACD$ addition property of equality
(subtract $\,m\angle BCD\,$ from both sides)
12.   [beautiful math coming... please be patient] $\,\angle DFB\cong \angle EFC$
vertical angles are congruent
13.   [beautiful math coming... please be patient] $\,\Delta BDF\cong \Delta CEF$
ASA (steps 8, 11, 12)
14.   [beautiful math coming... please be patient] $\,DF=EF\,$
(or, $\,\overline{DF} \cong \overline{EF}$ )
CPCTC

In the exercises, you will also practice doing shorter two-column 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.

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Is there an ‘SSA’ Congruence Theorem? No!


On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
(MAX is 10; there are 10 different problem types.)