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6.7 — Pythagoras by Proportions
1. In class: A Practice Understanding Task
Why the three right triangles are similar
Begin with a right triangle. Draw an altitude from its right-angle vertex to the hypotenuse—a line segment perpendicular to that side. The altitude divides the original triangle into two smaller right triangles.
Why are the three triangles similar? All three have a right angle. The triangle on the left shares angle L with the original triangle; the triangle on the right shares angle R. Two equal corresponding angles are enough to establish AA similarity. Consequently the original triangle and both smaller triangles are similar to one another.
The altitude divides the hypotenuse into pieces \(x\) and \(y\), so we also know \(x+y=c\). These facts are enough to discover the relationships we need.
Prove the Pythagorean theorem using proportions
In the large triangle, the left leg has length \(a\). In the smaller left triangle, the side next to the shared angle L has length \(x\), and its hypotenuse is \(a\). Matching the left small triangle to the large one gives
\[\frac{a}{c}=\frac{x}{a}\quad\Longrightarrow\quad a^2=cx.\]The same reasoning on the right gives
\[\frac{b}{c}=\frac{y}{b}\quad\Longrightarrow\quad b^2=cy.\]Add the two equations. Because the two hypotenuse pieces together are the whole hypotenuse, \(x+y=c\):
\[\begin{aligned}a^2+b^2&=cx+cy\\&=c(x+y)\\&=c\cdot c\\&=\boxed{c^2}.\end{aligned}\]Notice that we did not assume the Pythagorean theorem to derive it. The proof used angle similarity, equal side ratios, and the fact that the two hypotenuse pieces add to \(c\).
The altitude is the geometric mean of the two pieces
Compare the two smaller right triangles. Their sides correspond so that the altitude \(h\) in one triangle matches the segment \(y\) in the other, while \(x\) matches \(h\). Thus
\[\frac{x}{h}=\frac{h}{y}\quad\Longrightarrow\quad\boxed{h^2=xy}.\]The geometric mean of two positive numbers is the positive square root of their product: \(\sqrt{xy}\). Since a length is positive,
\[\boxed{h=\sqrt{xy}}.\]A right triangle's hypotenuse is divided into segments of lengths 9 and 16 by the altitude from the right angle. Find the altitude.
\[h^2=9\cdot16=144\quad\Longrightarrow\quad\boxed{h=12}.\]We take the positive square root because a length cannot be negative.
Each leg is a geometric mean, too
Recall the two proportions from the Pythagorean proof. They already give
\[\boxed{a^2=cx}\qquad\text{and}\qquad\boxed{b^2=cy}.\]So each leg is the geometric mean of the whole hypotenuse and its adjacent segment:
\[a=\sqrt{cx},\qquad b=\sqrt{cy}.\]Look at the diagram: the left leg \(a\) touches the left part \(x\) of the hypotenuse; the right leg \(b\) touches the right part \(y\). Pairing a leg with the wrong hypotenuse piece produces the wrong answer.
The whole hypotenuse is \(c=9+16=25\). Apply the adjacent-segment rule to each leg:
\[a^2=25(9)=225\Rightarrow \boxed{a=15},\] \[b^2=25(16)=400\Rightarrow \boxed{b=20}.\]Check with the Pythagorean theorem: \(15^2+20^2=225+400=625=25^2\).
Finding unknown lengths in altitude diagrams
Sometimes you know the altitude and one piece of the hypotenuse instead of both pieces. Find the other piece first, using \(h^2=xy\); then add the pieces to get \(c\). If you need a leg, use the full hypotenuse and its adjacent piece.
In an original example, the left piece of the hypotenuse is 4 and the altitude is 6. Find the other piece, the whole hypotenuse, and the left leg.
Step 1: \(6^2=4y\), so \(y=9\).
Step 2: \(c=4+9=13\).
Step 3: The left leg touches the piece of length 4, so
\[a^2=cx=13(4)=52\quad\Longrightarrow\quad \boxed{a=2\sqrt{13}}.\]Check: \(h^2=36=4\cdot 9\), as required.
2. Ready review — Recognize similar and congruent triangles
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Similar triangles have equal corresponding angles, with corresponding side lengths in the same ratio. Congruent triangles have equal corresponding angles and matching side lengths. Every pair of congruent triangles is also similar, but similar triangles can have different sizes.
| What the diagram tells you | What you can conclude |
|---|---|
| Two corresponding angles are marked equal (AA). | The triangles are similar. Their size need not match. |
| All three corresponding sides are marked equal (SSS). | The triangles are congruent. |
| Two corresponding angles and one corresponding side are equal (ASA/AAS). | The triangles are congruent. |
Two triangles both have angles of \(42^\circ\) and \(68^\circ\). Their third angles are \(70^\circ\), so they are similar by AA. If a corresponding side in the first is 5 and in the second is 10, they are not congruent. If those corresponding sides are instead marked equal, the equal scale factor makes the triangles congruent by AAS or ASA.
When reviewing the packet's triangles, start with the identical angle-arc patterns (A, C; B, G; E, F) and identical side-tick patterns (D, H). Decide congruence only when the marks justify it. Do not rely on which triangle looks bigger.
Check your work — Ready 1 answer key
- Similar group: A, B, C, E, F, G. The one-, two-, and three-arc angle markings identify the same three angle measures; the triangle sum supplies the unmarked third angle, so AA applies throughout. B and G are congruent by ASA (the marked included side matches); D and H are congruent by SSS and therefore also similar. No other congruence or similarity involving D/H is forced by the markings.
3. Set review — Use the right-triangle proportions
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Set problems 2–7 use the same picture, with the notation \(a,b,c\), altitude \(f\), and hypotenuse pieces \(d,e\). It is the same geometry as our diagram, but the symbols have changed: \(h\) becomes \(f\), \(x\) becomes \(d\), and \(y\) becomes \(e\).
Suppose you are asked to finish \(\dfrac{a}{c}=\dfrac{d}{\square}\). The left small triangle has hypotenuse \(a\), while the original has hypotenuse \(c\). Match the segment \(d\) in the small triangle to the side \(a\) in the large triangle:
\[\frac{a}{c}=\frac{d}{\boxed{a}}.\]Cross-multiplying yields \(a^2=cd\), the leg theorem.
A right triangle's altitude divides the hypotenuse into two pieces of lengths 4 and 25. Find the altitude and the original leg adjacent to the piece of length 4.
\[f^2=4\cdot25=100\quad\Longrightarrow\quad\boxed{f=10}.\]The whole hypotenuse is 29, so that original leg has squared length \(4\cdot29=116\), and therefore length \(\boxed{2\sqrt{29}}\).
A right triangle has legs 6 and 8, and an altitude from its right angle to the hypotenuse. Find the altitude.
First find the hypotenuse: \(c=\sqrt{6^2+8^2}=10\). The two hypotenuse pieces satisfy
\[x=\frac{a^2}{c}=\frac{36}{10}=\frac{18}{5},\qquad y=\frac{b^2}{c}=\frac{64}{10}=\frac{32}{5}.\]Then the altitude is
\[h=\sqrt{xy}=\sqrt{\frac{18}{5}\cdot\frac{32}{5}}=\boxed{\frac{24}{5}}.\]A right triangle has hypotenuse 13 and left leg 5. Find the altitude to the hypotenuse.
First find the other leg: \(b=\sqrt{13^2-5^2}=12\). Now use the leg relations to identify the two hypotenuse pieces:
\[x=\frac{5^2}{13}=\frac{25}{13},\qquad y=\frac{12^2}{13}=\frac{144}{13}.\]Finally,
\[h=\sqrt{xy}=\sqrt{\frac{25\cdot144}{169}}=\boxed{\frac{60}{13}}.\]For Set 8–9: if two original legs are given, first find the hypotenuse using the Pythagorean theorem; if a hypotenuse and one original leg are given, find the missing hypotenuse segment using the leg theorem. Once both hypotenuse pieces are known, the altitude follows from the altitude theorem. These are applications of the same relationships, not new trigonometry.
Check your work — Set 2–9 answer key
- \(\frac ac=\frac fb\): blank \(b\).
- \(\frac af=\frac cb\): blank \(b\).
- \(\frac ab=\frac fe\): blank \(e\).
- \(\frac ad=\frac ca\): blank \(a\).
- \(\frac fd=\frac ef\): blank \(f\).
- \(\frac bc=\frac eb\): blank \(b\).
- Hypotenuse \(13\); altitude \(\frac{60}{13}\), since \(h=\frac{5\cdot12}{13}\).
- Missing leg \(\sqrt{39}\); altitude \(\frac{5\sqrt{39}}8\). The hypotenuse is \(8\), so the missing leg squared is \(8^2-5^2=39\).
4. Go review — Solve proportions from parallel lines
A line segment parallel to one side of a triangle creates a smaller triangle with the same angles as the whole triangle. Corresponding side lengths are proportional. You may compare small to large lengths, or compare the upper portion to lower portion on each transversal; just keep the order consistent.
In an original diagram, a segment inside a triangle is parallel to one side. On the first transversal, the sections are 6 and 3. On the other, the corresponding sections are 10 and an unknown \(x\). Find \(x\).
\[\frac{6}{3}=\frac{10}{x}\quad\Longrightarrow\quad6x=30\quad\Longrightarrow\quad\boxed{x=5}.\]The same ratio holds on each transversal because the small and large triangles are similar.
Two sides of a triangle have total lengths 20 and 35. A parallel segment cuts off a smaller triangle with a corresponding side of length 8 on the first side. Find the lower section on the second side.
The small-to-large scale factor is \(8/20=2/5\). The corresponding small side on the second transversal is \((2/5)(35)=14\). Therefore the lower section is
\[35-14=\boxed{21}.\]Common mistake: the proportion \(8/20=z/35\) finds the whole upper section \(z\), not the leftover lower section asked for.
5. Go review — Complete arithmetic and geometric sequences
An arithmetic sequence changes by adding the same number each step, called the common difference. A geometric sequence changes by multiplying by the same number each step, called the common ratio.
| Type | How to move one term | Four sample terms |
|---|---|---|
| Arithmetic | Add 4 each step | 3, 7, 11, 15 |
| Geometric | Multiply by 3 each step | 2, 6, 18, 54 |
For a table, count the number of jumps between term numbers, not the number of term labels visible. The move from term 1 to term 4 contains three jumps.
Term 2 equals 11 and term 6 equals 27. Fill terms 3, 4, and 5.
Four jumps carry 11 to 27. The total change is \(27-11=16\); divide by four jumps to get a difference of 4.
\[\boxed{11,\ 15,\ 19,\ 23,\ 27}\qquad\text{(terms 2 through 6)}.\]Term 3 equals 64 and term 6 equals 8. Find terms 4 and 5, assuming the common ratio is real.
There are three jumps, so \(64r^3=8\), giving \(r^3=1/8\) and \(r=1/2\). Multiply by one-half at each jump:
\[\boxed{64,\ 32,\ 16,\ 8}\qquad\text{(terms 3 through 6)}.\]The common ratio is a multiplier, not a quantity to subtract. For positive values separated by three jumps, the real ratio here is uniquely \(1/2\).
Go — Check parallel-line proportions and sequences
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Use the parallel-line worked examples for problems 10–11 and the sequence worked examples for problems 12–15.
Check your work — Go 10–15 answer key
- Assuming the interior segment is parallel to the base, \(\frac{10}{5}=\frac{14}{x}\), so \(x=7\). Parallelism is not marked.
- Assuming the segment of length \(6\) is parallel to the outer side, \(\frac{5}{8}=\frac6x\), so the missing outer side is \(\frac{48}{5}=9.6\). Parallelism is not marked.
- Missing values: term 2 is \(12\), term 3 is \(17\); common difference \(5\).
- Missing values: term 2 is \(14\), term 3 is \(28\); common ratio \(2\).
- Missing values: term 6 is \(21\), term 7 is \(32\); common difference \(11\).
- Missing values: term 8 is \(6\), term 9 is \(12\); common ratio \(2\).
Additional Learning Resources
| Resource and direct link | What it helps with | Scope / What to skip |
|---|---|---|
| Khan Academy — Pythagorean theorem proof using similarity | A detailed demonstration of why an altitude creates three similar triangles, and how proportional legs prove \(a^2+b^2=c^2\). | Close match. This is the central proof in today's main task. |
| Khan Academy — Geometric mean theorem: finding a missing side (practice) | Practice identifying the right triangle's hypotenuse segments and solving for a missing side. | Close match. Uses the right-triangle geometric-mean relationships from the Main task and Set. |
| OpenStax — Contemporary Mathematics, 10.3 Triangles | A written review of triangle similarity, congruence, and related reasoning. | Some extra material. Use the congruence and similarity portions for Ready 1; skip unrelated applications and historical discussion. |
| Khan Academy — Arithmetic sequences review and Extend geometric sequences (practice) | Short reading and practice for the table-completion skills in Go 12–15. | Some extra material. Focus on common difference, common ratio, and extending terms; explicit/recursive sequence formulas are not needed here. |
These are optional online resources and require an internet connection. The lesson text and diagrams work offline; mathematical typesetting requires an internet connection.
Teacher review — Substantive packet issues
No substantive mathematical errors identified in Section 6.7. Its geometric-mean theorems and proof directions are consistent with the diagrams. The main task relies on students matching corresponding sides after rotating or reflecting cutouts; the proportional relationships are supplied explicitly above.
The Set diagrams rely on the heading's statement that the figures are right triangles with an altitude to the hypotenuse. Reinforce this given when discussing the diagrams, even where a right-angle marker is visually small.
Section 6.7 packet-to-lesson coverage
| Curriculum component | Original problems / objectives | Where this lesson teaches it |
|---|---|---|
| Main task (pp. 37–38) | Steps 1–6: three similar triangles and proportions | In class |
| Main task (p. 37) | Steps 7–8: prove the Pythagorean theorem | In class |
| Main task (p. 38) | Altitude theorem 1, altitude theorem 2, final length problem | In class |
| Ready (p. 39) | 1: AA similarity, triangle congruence via marked sides/angles | Section 2 |
| Set (p. 40) | 2–7: complete ratios; 8–9: solve right-triangle altitude diagrams | Section 3 |
| Go (p. 41) | 10–11: nested parallel-line similar triangles | Section 4 + Go practice A–B |
| Go (p. 41) | 12–15: arithmetic and geometric sequence table completion | Section 5 + Go practice C–D |
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