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Pythagorean Theorem


Alignments to Content Standards: G-SRT.B.4

Task

Below is a picture of a right triangle $ABC$ with right angle $C$ along with the point $D$ so that $\overleftrightarrow{CD}$ is perpendicular to $\overleftrightarrow{AB}$.

Pythagoras1_be33ab9a0a6d5b98b49483742582ac33

  1. Show that $\triangle ACB$ is similar to $\triangle ADC$ and to $\triangle CDB$.
  2. Use part (a) to conclude that $|AC|^2 + |BC|^2 = |AB|^2$.

IM Commentary

The purpose of this task is to prove the Pythagorean theorem using similar triangles. The teacher may wish to be slightly less leading in part (a) by suggesting for students to examine possible similarities between the three triangles without giving the precise correspondences. On the other hand, in part (b), the teacher may wish to provide more guidance as there are many proportional relationships which follow from the similarities established in part (a) and it takes quite a bit of insight or trial and error to find the ones which produce the Pythagorean Theorem.

The key insight required to solve part (b) is to identify that each of the terms $|AC|^2$ and $|BC|^2$ in the Pythagorean Theorem can come naturally via one and only one proportion because, for example, the side $|AC|$ appears in both $\triangle ACB$ and$\triangle ADC$. This means that we can get an $|AC|^2$ term by comparing $|AC|$ and $|AB|$ on $\triangle ACB$ to $|AD|$ and $|AC|$ on $\triangle ADC$. It is not possible to get $|AB|^2$ in this way because $\overline{AB}$ only occurs in $\triangle ACB$: therefore this piece occurs later as a byproduct of algebraic manipulation.

In addition to providing an interesting application of similarity of triangles, this thought process about which proportions to use is the most valuable aspect of this task. It exemplifies MP7, Look For and Make Use of Structure. Trial and error will eventually lead to seeing the Pythagorean Theorem but careful thought about the similar triangles and the relationship to be shown can guide an informed strategy.

Solution

  1. We first show that $\triangle ACB$ is similar to $\triangle ADC$. We know that

    • $m(\angle ACB) = m(\angle CDA)$ by hypothesis,
    • $m(\angle CAB) = m(\angle DAC)$ because these are the same angle.

    Therefore by AA similarity, $\triangle ACB$ is similar to $\triangle ADC$. Alternatively $\angle DCA$ is complementary to $\angle DAC$ and $\angle CBA$ is complementary to $\angle CAB$. Angles complementary to congruent angles must be congruent so $\angle DCA$ is congruent to $\angle CBA$. By AAA similiarty $\triangle ACB$ is similar to $\triangle ADC$.

    Similar reasoning will establish that $\triangle ACB$ is similar to $\triangle CDB$:

    • $m(\angle ACB) = m(\angle CDB)$ by hypothesis,
    • $m(\angle CBA) = m(\angle DBC)$ because these are the same angle.

    By AA similarity (or AAA similarity if we make an extra argument as above), $\triangle ACB$ is similar to $\triangle CDB$.

  2. The two triangle similarity results from part (a) allow us to set up different proportions involving corresponding pairs of sides in the similar triangles. With two pairs of similar triangles (three if we include the fact that $\triangle ADC$ is similar to $\triangle DCB$ since both are similar to $\triangle ACB$) this gives us six possible proportions. Some care is needed in order to choose the ones which will give the Pythagorean theorem without too much trial and error.

    Suppose we set $a = |BC|$, $b = |AC|$, $c = |AB|$. In order to get an $a^2$ we would like to use the two occurrences of $\overline{BC}$, one in $\triangle ACB$ and one in $\triangle CDB$: $$ \frac{|BC|}{|BA|} = \frac{|BD|}{|BC|} $$ which we can rewrite as $$ \frac{a}{c} = \frac{|BD|}{a}. $$ Similarly, to get a $b^2$ we need to use the two occurrences of $\overline{AC}$, one in $\triangle ACB$ and the other in $\triangle ADC$: $$ \frac{|AC|}{|AB|} = \frac{|AD|}{|AC|} $$ which we can rewrite as $$ \frac{b}{c} = \frac{|AD|}{b}. $$

    Clearing denominators in our two fraction equivalences gives us $a^2 = c|BD|$ and $b^2 = c|AD|$. Adding these two equations together gives

    \begin{align} a^2 + b^2 &= c(|BD| + |AD|) \\ &= c|AB| \\ &= c^2. \end{align}

    This is the Pythagorean Theorem.