Geometry · 8.G.B.6 · G.SRT.B.4

Proving the Pythagorean theorem with similar triangles

One segment dropped from the square corner cuts a right triangle into two smaller copies of itself. How do those copies prove the Pythagorean theorem?

You need

Geometric means in a right triangleGeometry Proving two triangles similarGeometry

Where you are 4 in Proofs about similar figures

Geometry11

Geometric proof18

Proofs about similar figures4

Proving the Pythagorean theorem by rearrangement Proving two triangles similar

Proving the Pythagorean theorem with similar triangles

Proving the triangle proportionality theorem

When you reach for it

The Pythagorean theorem says that in a right triangle with legs a and b and hypotenuse c,

a^2 + b^2 = c^2

The rearrangement proof gets there by counting areas. This proof uses no area at all. It uses one segment and the idea of proving two triangles similar: drop the altitude from the right angle to the hypotenuse, notice that it cuts the triangle into two smaller triangles with the same angles as the whole, and read proportions off those similarities. Two of the proportions, added together, are the theorem.

Reach for it when a proof about a right triangle has to be made with lengths and ratios, when a figure shows a right triangle with the altitude to its hypotenuse drawn in, and whenever a problem gives you some of the six lengths in that figure (two legs, the hypotenuse, the altitude and the two pieces of the hypotenuse) and asks for the rest. Along the way the proof gives two more facts about that figure that are worth knowing for their own sake.

The procedure

The figure and its names. Start with \triangle ABC with its right angle at C, so that \overline{AB} is the hypotenuse. Draw the altitude from C: the segment from C perpendicular to \overline{AB}, meeting it at a point D, called the foot of the altitude. Name the six lengths as in Figure 1:

a = BC \qquad b = AC \qquad c = AB

h = CD \qquad p = AD \qquad q = DB

Figure 1The altitude \overline{CD} splits the hypotenuse into p and q.

Each piece of the hypotenuse touches one leg: p touches b at A, and q touches a at B.

First, the foot lies between A and B. The proof will need p + q = c, and that is only true if D lands on the hypotenuse between its ends. The figure shows it, but a figure may show where things are without proving it (what a diagram may assume), so it gets a reason.

The angles of \triangle ABC add to 180^\circ (the triangle angle sum), and the right angle takes 90^\circ, so \angle A and \angle B add to 90^\circ and each is acute.

D is not B. If it were, \overline{CB} would be perpendicular to \overline{AB}, and the triangle would have right angles at both B and C, adding to 180^\circ before \angle A is counted. In the same way D is not A.
D is not beyond B. If it were, \triangle CDB would have a right angle at D, and its angle at B would be the supplement of the acute \angle ABC, so obtuse. A triangle cannot hold a right angle and an obtuse angle together. In the same way D is not beyond A.

That leaves one place: D lies strictly between A and B. By the segment addition postulate, p + q = c.

Second, the three similar triangles. The altitude makes two smaller right triangles, \triangle ADC and \triangle CDB. Each shares an acute angle with the whole triangle.

\triangle ADC and \triangle ACB both have a right angle, at D and at C, and both contain \angle A, because D lies on \overline{AB}. Two pairs of angles: AA similarity gives \triangle ADC \sim \triangle ACB, with A \leftrightarrow A, D \leftrightarrow C, C \leftrightarrow B.
\triangle CDB and \triangle ACB both have a right angle, at D and at C, and both contain \angle B. By AA, \triangle CDB \sim \triangle ACB, with C \leftrightarrow A, D \leftrightarrow C, B \leftrightarrow B.

The second similarity says more than it first seems to. Its correspondence pairs the angle of \triangle CDB at C with the angle of \triangle ACB at A, so \angle DCB \cong \angle A: the altitude splits the right angle at C into a copy of \angle A and a copy of \angle B. The two small triangles then have right angles at D and the equal angles \angle DAC and \angle DCB, so by AA once more they are similar to each other.

All three facts fit in one line, with the letters in matching positions:

\triangle ACB \sim \triangle ADC \sim \triangle CDB

Read it by positions. Position 1 is the angle equal to \angle A in each triangle (A, A, C); position 2 is the right angle (C, D, D); position 3 is the angle equal to \angle B (B, C, B). Figure 2 turns the three triangles to face the same way so the matching is visible.

Figure 2The three triangles pulled apart and turned to face the same way. Matching corners carry matching marks.

Third, the proportions. A side is named by two positions, so the partners are:

Side	\triangle ACB	\triangle ADC	\triangle CDB
positions 1 and 2	b	p	h
positions 2 and 3	a	h	q
positions 1 and 3	c	b	a

Corresponding sides of similar triangles are proportional, so for any two columns, the ratio of the entries is the same in all three rows. Comparing the columns two at a time gives three chains:

\begin{aligned} \triangle ACB \text{ and } \triangle ADC&: \quad \frac{b}{p} = \frac{a}{h} = \frac{c}{b} \\ \triangle ACB \text{ and } \triangle CDB&: \quad \frac{b}{h} = \frac{a}{q} = \frac{c}{a} \\ \triangle ADC \text{ and } \triangle CDB&: \quad \frac{p}{h} = \frac{h}{q} = \frac{b}{a} \end{aligned}

From the first chain, \displaystyle \frac{c}{b} = \frac{b}{p}, and clearing the fractions gives b^2 = cp. From the second, \displaystyle \frac{a}{q} = \frac{c}{a} gives a^2 = cq. Add them, and use p + q = c:

\begin{aligned} a^2 + b^2 &= cq + cp \\ &= c(p + q) \\ &= c \cdot c = c^2 \end{aligned}

That is the whole proof. Here it is in two columns, with the lengths written by their endpoints.

Given. \triangle ABC has a right angle at C; \overline{CD} \perp \overline{AB} with D on line AB.

Prove. AC^2 + BC^2 = AB^2.

Statements	Reasons
1. \angle ACB is a right angle; \overline{CD} \perp \overline{AB}	given
2. D lies between A and B	\angle A and \angle B are acute, so no other place is possible (shown above): line 1
3. \angle ADC and \angle CDB are right angles	definition of perpendicular lines: line 1
4. \angle ADC \cong \angle ACB	Right Angle Congruence Theorem: lines 1 and 3
5. \angle DAC \cong \angle CAB	reflexive property of congruence: both are \angle A, since D lies on \overline{AB} (line 2)
6. \triangle ADC \sim \triangle ACB	AA similarity: lines 4 and 5
7. \displaystyle {\frac{AD}{AC} = \frac{AC}{AB}}	corresponding sides of similar triangles are proportional: line 6
8. {AC^2 = AB \cdot AD}	multiplication property of equality: line 7
9. \angle CDB \cong \angle ACB	Right Angle Congruence Theorem: lines 1 and 3
10. \angle DBC \cong \angle CBA	reflexive property of congruence: both are \angle B
11. \triangle CDB \sim \triangle ACB	AA similarity: lines 9 and 10
12. \displaystyle {\frac{DB}{CB} = \frac{CB}{AB}}	corresponding sides of similar triangles are proportional: line 11
13. {BC^2 = AB \cdot DB}	multiplication property of equality: line 12
14. {AC^2 + BC^2 = AB(AD + DB)}	addition property of equality and the distributive property: lines 8 and 13
15. {AD + DB = AB}	segment addition postulate: line 2
16. {AC^2 + BC^2 = AB^2}	substitution: line 15 into line 14

Lines 8 and 13 are b^2 = cp and a^2 = cq, with endpoints in place of letters. The third similarity was not needed for the theorem.

Two facts on the way: geometric means. The geometric mean of two positive numbers x and y is the positive number g with

\frac{x}{g} = \frac{g}{y}

that is, g^2 = xy and \displaystyle g = \sqrt{xy}. The geometric mean of 4 and 9 is 6, since \displaystyle \frac{4}{6} = \frac{6}{9}. The proof has already found two of them, and the third chain gives another.

Each leg is the geometric mean of the hypotenuse and the piece touching that leg: b^2 = cp and a^2 = cq (geometric mean and a leg).
The altitude is the geometric mean of the two pieces: from \displaystyle \frac{p}{h} = \frac{h}{q} in the third chain, which compares the two small triangles, h^2 = pq (geometric mean and the altitude).

The first chain holds one more useful fact, \displaystyle \frac{a}{h} = \frac{c}{b}, which clears to ab = ch. Both sides are twice the area of \triangle ABC, counted once with the legs as base and height and once with the hypotenuse and the altitude (area of a triangle).

The same proof, told with areas of similar figures. Similar figures have areas in the ratio of the squares of their corresponding lengths (scale factor and area). The three similar triangles in Figure 2 have hypotenuses c, b and a, so their areas are kc^2, kb^2 and ka^2 for one and the same number k. The two small triangles fill the whole one exactly, so ka^2 + kb^2 = kc^2, and dividing by k gives the theorem again. Run the other way, the theorem works for any shape: three similar figures standing on a, b and c, with those sides in corresponding places, have areas ma^2, mb^2 and mc^2 for one number m, and multiplying a^2 + b^2 = c^2 by m says the figure on the hypotenuse equals the two on the legs together. Squares are only the most familiar case. Example 5 uses half circles.

Worked examples

Example 1

From the two pieces to everything

In right \triangle ABC with the right angle at C, the altitude \overline{CD} cuts the hypotenuse into AD = 4 and DB = 9. Find CD, AC and BC, and check the Pythagorean theorem on the answers.

Solution

The two pieces are p = 4 and q = 9, and the whole hypotenuse is c = 4 + 9 = 13, since D lies between A and B. The altitude is the geometric mean of the pieces:

h^2 = pq = 4 \cdot 9 = 36 \qquad\Longrightarrow\qquad CD = 6

Each leg is the geometric mean of the hypotenuse and the piece touching it. \overline{AC} touches \overline{AD} at A, and \overline{BC} touches \overline{DB} at B:

\begin{aligned} AC^2 &= 13 \cdot 4 = 52 \\ BC^2 &= 13 \cdot 9 = 117 \end{aligned}

so \displaystyle AC = \sqrt{52} = 2\sqrt{13} and \displaystyle BC = \sqrt{117} = 3\sqrt{13}.

Check. AC^2 + BC^2 = 52 + 117 = 169 = 13^2, as the theorem says. The area agrees as well: \displaystyle AC \cdot BC = 2\sqrt{13} \cdot 3\sqrt{13} = 6 \cdot 13 = 78, and AB \cdot CD = 13 \cdot 6 = 78. Figure 1 is drawn to these numbers.

Example 2

From the legs to the pieces

A right triangle has legs AC = 20 and BC = 21. Find the two pieces into which the altitude from C cuts the hypotenuse, and the altitude itself.

Solution

The theorem, now proved, gives the hypotenuse: AB^2 = 20^2 + 21^2 = 400 + 441 = 841, so AB = 29. Each leg relation, solved for the piece, gives that piece:

\begin{aligned} AD &= \frac{AC^2}{AB} = \frac{400}{29} \\ DB &= \frac{BC^2}{AB} = \frac{441}{29} \end{aligned}

The pieces are about 13.79 and 15.21. They add to \displaystyle \frac{841}{29} = 29, the whole hypotenuse, as they must.

For the altitude, ab = ch is the quickest route:

CD = \frac{AC \cdot BC}{AB} = \frac{20 \cdot 21}{29} = \frac{420}{29} \approx 14.48

Check. The altitude relation gives the same number: \displaystyle AD \cdot DB = \frac{400 \cdot 441}{29^2}, whose square root is \displaystyle \frac{20 \cdot 21}{29}.

Example 3

The altitude and the hypotenuse

The altitude to the hypotenuse of a right triangle is 6, and the hypotenuse is 20. Find the two pieces of the hypotenuse and the two legs.

Solution

Call the piece at A x. The pieces add to the hypotenuse, so the piece at B is 20 - x, and the altitude relation h^2 = pq becomes an equation in x:

\begin{aligned} x(20 - x) &= 36 \\ x^2 - 20x + 36 &= 0 \\ (x - 2)(x - 18) &= 0 \end{aligned}

So x = 2 or x = 18. Both are the same triangle: the two answers only swap which end of the hypotenuse is called A. Take AD = 2 and DB = 18. Then

\begin{aligned} AC^2 &= 20 \cdot 2 = 40, &\quad AC &= 2\sqrt{10} \\ BC^2 &= 20 \cdot 18 = 360, &\quad BC &= 6\sqrt{10} \end{aligned}

Check. 40 + 360 = 400 = 20^2.

Not every pair of numbers works. With altitude 6 and hypotenuse 10 the same steps give x^2 - 10x + 36 = 0, whose discriminant 100 - 144 is negative, so there is no such right triangle. Example 6 shows why: the altitude can never be more than half the hypotenuse.

Example 4

The three relations in coordinates

In Figure 3, A(0, 0), B(6, 8) and C(8, 4). Show that \triangle ABC has a right angle at C, find the foot D of the altitude from C, and check the three relations.

Figure 3A(0, 0), B(6, 8), C(8, 4), and the foot D of the altitude from C.

Solution

The right angle. The slope of \overline{CA} is \displaystyle \frac{4 - 0}{8 - 0} = \frac{1}{2} and the slope of \overline{CB} is \displaystyle \frac{8 - 4}{6 - 8} = -2. Their product is -1, so the legs are perpendicular (perpendicular slopes), and the right angle is at C.

The lengths. By the distance formula,

\begin{aligned} AC^2 &= 8^2 + 4^2 = 80 \\ BC^2 &= 2^2 + 4^2 = 20 \\ AB^2 &= 6^2 + 8^2 = 100 \end{aligned}

so AB = 10.

The foot. The leg relation gives the piece at A without any equation of a line:

AD = \frac{AC^2}{AB} = \frac{80}{10} = 8

and then DB = 10 - 8 = 2. So D is \displaystyle \frac{8}{10} = \frac{4}{5} of the way from A to B, and since A is the origin, its coordinates are \displaystyle \frac{4}{5} of those of B:

D = \left(\tfrac{4}{5} \cdot 6,\ \tfrac{4}{5} \cdot 8\right) = (4.8,\ 6.4)

The checks. The slope of \overline{CD} is \displaystyle \frac{6.4 - 4}{4.8 - 8} = -\frac{3}{4}, and the slope of \overline{AB} is \displaystyle \frac{8}{6} = \frac{4}{3}; the product is -1, so \overline{CD} really is perpendicular to \overline{AB}. The distance formula gives CD^2 = 3.2^2 + 2.4^2 = 10.24 + 5.76 = 16, and the three relations hold:

Relation	Left side	Right side
CD^2 = AD \cdot DB	16	8 \cdot 2 = 16
AC^2 = AB \cdot AD	80	10 \cdot 8 = 80
BC^2 = AB \cdot DB	20	10 \cdot 2 = 20

The distance formula is itself built on the Pythagorean theorem, so this is a check on one triangle, not a second proof.

Example 5

Half circles on the sides

A right triangle has legs 24 and 32. Half circles are drawn outward on all three sides, each with its side as the diameter, as in Figure 4. Show that the half circle on the hypotenuse has the same area as the other two together. Then show the same for the two small triangles made by the altitude.

Figure 4Half circles on the three sides of a right triangle.

Solution

The hypotenuse is \displaystyle \sqrt{24^2 + 32^2} = \sqrt{1600} = 40. A half circle on a diameter d has radius \displaystyle \frac{d}{2}, so its area is \displaystyle \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8} (area of a circle):

\begin{aligned} \text{on the leg } 24&: \ \frac{576\pi}{8} = 72\pi \\ \text{on the leg } 32&: \ \frac{1024\pi}{8} = 128\pi \\ \text{on the hypotenuse}&: \ \frac{1600\pi}{8} = 200\pi \end{aligned}

and 72\pi + 128\pi = 200\pi. Nothing about half circles mattered except that all half circles are similar, so each area is one fixed number, \displaystyle \frac{\pi}{8}, times its side squared. Multiplying 24^2 + 32^2 = 40^2 by that number is the whole reason.

The triangles. The whole triangle has area \displaystyle \frac{1}{2} \cdot 24 \cdot 32 = 384. The small triangle standing on the leg 24 as its hypotenuse is similar to the whole with scale factor \displaystyle \frac{24}{40} = 0.6, so its area is 0.6^2 \cdot 384 = 138.24; the one on the leg 32 has scale factor 0.8 and area 0.8^2 \cdot 384 = 245.76. Together:

138.24 + 245.76 = 384

which had to happen, since the two small triangles fill the whole one. Read backwards, this is the area version of the proof in the procedure. The general statement, that the figure on the hypotenuse equals the similar figures on the legs, is Proposition 31 of Book VI of Euclid’s Elements.

Example 6

A geometric mean with compass and straightedge

Two segments have lengths p and q. Lay them end to end as \overline{PQ} and \overline{QR}, draw the half circle on \overline{PR}, and raise a perpendicular at Q to meet it at T, as in Figure 5. Prove that \displaystyle QT = \sqrt{pq} and that \displaystyle \sqrt{pq} \le \frac{p + q}{2}. Then use it to draw a segment of length \displaystyle \sqrt{5} from a segment of length 1.

Figure 5The height \overline{QT} is the geometric mean of p and q; the radius \overline{OT} is their average.

Solution

The right triangle. T lies on the half circle whose diameter is \overline{PR}, so \angle PTR is a right angle (an angle in a half circle is right). In the right triangle PTR, the segment \overline{TQ} runs from the right angle perpendicular to the hypotenuse \overline{PR}: it is the altitude, and its foot Q cuts the hypotenuse into p and q. The altitude relation gives

QT^2 = PQ \cdot QR = pq

so \displaystyle QT = \sqrt{pq}.

The inequality. The center O of the half circle is the midpoint of \overline{PR}, so the radius is \displaystyle OT = \frac{p + q}{2}, the ordinary average of p and q. If Q is not O, then \triangle OQT has a right angle at Q, and \overline{OT} is its hypotenuse, so it is longer than the leg \overline{QT} (by the theorem just proved, OT^2 = OQ^2 + QT^2 > QT^2). If Q is O, the two are equal. So

\sqrt{pq} \le \frac{p + q}{2}

with equality exactly when Q is the center, that is, when p = q. The geometric mean of two positive numbers is never more than their average. This is also the reason Example 3 had a limit: the altitude \displaystyle \sqrt{pq} can be at most \displaystyle \frac{p + q}{2}, half the hypotenuse.

Drawing \displaystyle \sqrt{5}. Take p = 1 and q = 5, so PR = 6 and the radius is 3. The perpendicular at Q meets the half circle at a height \displaystyle \sqrt{1 \cdot 5} = \sqrt{5} \approx 2.24, less than the radius 3 as the inequality requires. Compass and straightedge alone have drawn an exact irrational length.

Boundaries

The angle at C must be right. Every similarity in the proof pairs the right angle at D with a right angle at C. In a triangle with no right angle, the altitude from C still makes right triangles, but neither is similar to the whole, and AC^2 = AB \cdot AD fails (Exercise 16).

The foot must lie between A and B. The last step, c(p + q) = c^2, needs p + q = c. In a right triangle the altitude from the right angle always lands strictly inside the hypotenuse, which is why the procedure proves it first. The altitude from an acute vertex of an obtuse triangle lands outside the opposite side, and there the pieces do not add up to the side.

The altitude is at most half the hypotenuse. Six lengths in the figure are tied by these relations, and two of them fix the rest, but not every two numbers are allowed: an altitude longer than half the hypotenuse belongs to no right triangle (Examples 3 and 6).

Coordinates check, they do not prove. The distance formula already uses the Pythagorean theorem, so a coordinate computation like Example 4 confirms the relations on one triangle and proves nothing about all of them.

It proves one direction. The proof starts from a right angle. The statement that a triangle whose sides satisfy a^2 + b^2 = c^2 has a right angle is the converse of the Pythagorean theorem, with its own proof. Exercise 17 proves a converse of the altitude relation.

Where this goes next

The two geometric mean facts have their own lessons, geometric mean and a leg and geometric mean and the altitude, and together with the theorem they turn any two of the six lengths in the figure into all six. The same pattern, a similarity read as a product of lengths, proves the power of a point: the circle on \overline{AB} as diameter passes through C, and h^2 = pq is the statement DA \cdot DB = DC^2 for the point D inside it. The sibling proof of the triangle proportionality theorem uses a parallel line where this one uses an altitude.

The first chain of ratios, \displaystyle \frac{c}{b} = \frac{b}{p}, says that \displaystyle \frac{p}{b} and \displaystyle \frac{b}{c} are the same number. Both are ratios in right triangles that contain \angle A, and that shared ratio is the cosine of \angle A (right triangle trigonometry). Dropping the same kind of altitude in a triangle with no right angle is how the law of cosines is derived, and it contains this theorem as the case of a right angle. The areas of similar figures, used in Example 5, come back in the volumes of similar solids, where the square becomes a cube.

Exercises

Reading the three similarities

Right \triangle JKL has its right angle at L, and \overline{LM} is the altitude to the hypotenuse, with M on \overline{JK}. Write three similarity statements: the whole \triangle JLK with each smaller triangle, and the two smaller triangles with each other, with the letters in matching positions.
In the figure of Exercise 1, write the proportion that gives LK^2 = MK \cdot JK and the one that gives LM^2 = JM \cdot MK, and name the similarity each comes from.

Finding lengths

In Exercises 3 to 10, \triangle ABC has a right angle at C, and \overline{CD} is the altitude to the hypotenuse \overline{AB}, as in Figure 1. Give exact answers.

AD = 5 and DB = 20. Find CD, AC and BC.
AD = 8 and DB = 10. Find CD, AC and BC.
AC = 12 and AB = 16. Find AD, DB, CD and BC.
AC = 9 and BC = 40. Find AB, AD, DB and CD.
CD = 10 and AD = 4. Find DB, AB, AC and BC.
BC = 10 and DB = 4. Find AB, AD, CD and AC.
CD = 12, and DB is 10 longer than AD. Find AD, DB, AC and BC.
CD = 5 and AB = 10. Find AD and DB, and say what kind of right triangle \triangle ABC is. Then explain why no right triangle has CD = 6 and AB = 10.

In coordinates

A(-1, 0), B(9, 0) and C(0, 3). Show that the right angle of \triangle ABC is at C, name the foot D of the altitude from C, and check all three relations.
A(0, 0), B(10, 5) and C(6, 8). Show that the right angle is at C, use \displaystyle AD = \frac{AC^2}{AB} to find the coordinates of the foot D of the altitude from C, and check that CD^2 = AD \cdot DB.
C(0, 0), A(6, 0) and B(0, 3). Find the coordinates of the foot D of the altitude from C and the length CD.

Similar figures on the sides

A right triangle has legs 3 and 5. Equilateral triangles are built outward on all three sides. An equilateral triangle with side s has area \displaystyle \frac{\sqrt{3}}{4}s^2. Find the three areas and check that the largest is the sum of the other two.
Similar figures stand on the two legs of a right triangle, with areas 20\ \text{cm}^2 and 45\ \text{cm}^2, and a third similar figure stands on the hypotenuse, placed the same way. Find its area and the ratio of the shorter leg to the longer.

Reasoning

In \triangle ABC the angle at C is obtuse, and \overline{CD} is the altitude to \overline{AB}. Which line of the two-column proof in the procedure is false now, and why?
D lies between A and B, \overline{CD} \perp \overline{AB}, and CD^2 = AD \cdot DB. Prove that \angle ACB is a right angle.
\triangle ABC has a right angle at C, and M is the midpoint of \overline{AB}. The line perpendicular to \overline{AB} at M meets ray BC at P and ray AC at Q. Prove that 4 \cdot MP \cdot MQ = AB^2.
Prove \triangle ADC \sim \triangle CDB directly by AA, without using the whole triangle \triangle ACB as a go-between.
Show that in every right triangle \displaystyle \frac{1}{h^2} = \frac{1}{a^2} + \frac{1}{b^2}, where h is the altitude to the hypotenuse. Check it on Example 1.
Show that the two pieces of the hypotenuse are in the ratio of the squares of the legs: \displaystyle \frac{p}{q} = \frac{b^2}{a^2}.
Name the lines of the two-column proof in the procedure that use the right angle at C, and say what each one would lose without it.

Answers

\triangle JLK \sim \triangle JML, \triangle JLK \sim \triangle LMK and \triangle JML \sim \triangle LMK. Position 1 holds the angle equal to \angle J (J, J, L), position 2 the right angle (L, M, M), position 3 the angle equal to \angle K (K, L, K). The first two are AA with a right angle and a shared acute angle; the third follows from them.
From \triangle JLK \sim \triangle LMK: \displaystyle \frac{LK}{MK} = \frac{JK}{LK}, pairing positions 2 and 3, then 1 and 3; clearing fractions gives LK^2 = MK \cdot JK. From \triangle JML \sim \triangle LMK: \displaystyle \frac{JM}{LM} = \frac{ML}{MK}, pairing positions 1 and 2, then 2 and 3; clearing fractions gives LM^2 = JM \cdot MK.
CD = 10, \displaystyle AC = 5\sqrt{5}, \displaystyle BC = 10\sqrt{5}. AB = 25; CD^2 = 5 \cdot 20 = 100; AC^2 = 25 \cdot 5 = 125; BC^2 = 25 \cdot 20 = 500. Check: 125 + 500 = 625 = 25^2.
\displaystyle CD = 4\sqrt{5}, AC = 12, \displaystyle BC = 6\sqrt{5}. AB = 18; CD^2 = 8 \cdot 10 = 80; AC^2 = 18 \cdot 8 = 144; BC^2 = 18 \cdot 10 = 180. Check: 144 + 180 = 324 = 18^2.
AD = 9, DB = 7, \displaystyle CD = 3\sqrt{7}, \displaystyle BC = 4\sqrt{7}. \displaystyle AD = \frac{AC^2}{AB} = \frac{144}{16}; DB = 16 - 9; CD^2 = 9 \cdot 7 = 63; BC^2 = 16 \cdot 7 = 112. Check: 144 + 112 = 256 = 16^2.
AB = 41, \displaystyle AD = \frac{81}{41}, \displaystyle DB = \frac{1600}{41}, \displaystyle CD = \frac{360}{41}. AB^2 = 81 + 1600 = 1681; each piece is its leg squared over 41; \displaystyle CD = \frac{9 \cdot 40}{41}. The pieces add to \displaystyle \frac{1681}{41} = 41.
DB = 25, AB = 29, \displaystyle AC = 2\sqrt{29}, \displaystyle BC = 5\sqrt{29}. \displaystyle DB = \frac{CD^2}{AD} = \frac{100}{4}; AC^2 = 29 \cdot 4 = 116; BC^2 = 29 \cdot 25 = 725. Check: 116 + 725 = 841 = 29^2.
AB = 25, AD = 21, \displaystyle CD = 2\sqrt{21}, \displaystyle AC = 5\sqrt{21}. \displaystyle AB = \frac{BC^2}{DB} = \frac{100}{4}; AD = 25 - 4; CD^2 = 21 \cdot 4 = 84; AC^2 = 25 \cdot 21 = 525. Check: 525 + 100 = 625.
AD = 8, DB = 18, \displaystyle AC = 4\sqrt{13}, \displaystyle BC = 6\sqrt{13}. With AD = x: x(x + 10) = 144, so x^2 + 10x - 144 = 0 and (x + 18)(x - 8) = 0; a length is positive, so x = 8. Then AB = 26, AC^2 = 26 \cdot 8 = 208 and BC^2 = 26 \cdot 18 = 468. Check: 208 + 468 = 676 = 26^2.
AD = DB = 5, and the triangle is isosceles, with legs \displaystyle 5\sqrt{2}. From x(10 - x) = 25, x^2 - 10x + 25 = (x - 5)^2 = 0. With CD = 6 the equation is x^2 - 10x + 36 = 0, whose discriminant 100 - 144 = -44 is negative, so no length x works: the altitude would be more than half the hypotenuse, which Example 6 rules out.
The slope of \overline{CA} is \displaystyle \frac{0 - 3}{-1 - 0} = 3 and of \overline{CB} is \displaystyle \frac{0 - 3}{9 - 0} = -\frac{1}{3}; the product is -1, so the right angle is at C. AB lies on the x-axis, so the foot is D(0, 0), with AD = 1, DB = 9, CD = 3, AB = 10. Then CD^2 = 9 = 1 \cdot 9, AC^2 = 1 + 9 = 10 = 10 \cdot 1 and BC^2 = 81 + 9 = 90 = 10 \cdot 9.
D(8, 4). Slopes: \overline{CA} has \displaystyle \frac{8}{6} = \frac{4}{3}, \overline{CB} has \displaystyle \frac{5 - 8}{10 - 6} = -\frac{3}{4}, product -1. AC^2 = 100, BC^2 = 25, AB^2 = 125, so \displaystyle AD = \frac{100}{\sqrt{125}} = 4\sqrt{5}, which is \displaystyle \frac{100}{125} = \frac{4}{5} of AB, and \displaystyle D = \frac{4}{5}(10, 5) = (8, 4). Then CD^2 = 2^2 + 4^2 = 20, and \displaystyle DB = \frac{25}{\sqrt{125}} = \sqrt{5}, so \displaystyle AD \cdot DB = 4\sqrt{5} \cdot \sqrt{5} = 20.
D(1.2, 2.4) and \displaystyle CD = \frac{6\sqrt{5}}{5} \approx 2.68. AC^2 = 36, AB^2 = 45, so D is \displaystyle \frac{AD}{AB} = \frac{AC^2}{AB^2} = \frac{36}{45} = \frac{4}{5} of the way from A to B: \displaystyle D = (6 - \frac{4}{5} \cdot 6,\ \frac{4}{5} \cdot 3) = (1.2, 2.4). Then \displaystyle CD^2 = 1.44 + 5.76 = 7.2 = \frac{36}{5}. Check: \displaystyle CD = \frac{AC \cdot BC}{AB} = \frac{18}{3\sqrt{5}}.
\displaystyle \frac{9\sqrt{3}}{4}, \displaystyle \frac{25\sqrt{3}}{4} and \displaystyle \frac{17\sqrt{3}}{2}. The hypotenuse squared is 9 + 25 = 34, so the largest is \displaystyle \frac{34\sqrt{3}}{4} = \frac{17\sqrt{3}}{2}, and \displaystyle \frac{9\sqrt{3}}{4} + \frac{25\sqrt{3}}{4} = \frac{34\sqrt{3}}{4}.
65\ \text{cm}^2, and the ratio is \displaystyle \frac{2}{3}. The areas are ka^2, kb^2 and kc^2 for one number k, so the third is 20 + 45. The legs are in the ratio \displaystyle \sqrt{\frac{20}{45}} = \sqrt{\frac{4}{9}} = \frac{2}{3}.
Line 4, \angle ADC \cong \angle ACB. The angle at D is still right, but \angle ACB is now obtuse, and no angle of \triangle ACB is right, so the right angle at D has no partner and \triangle ADC is not similar to \triangle ACB. Line 9 fails the same way. The conclusion fails too: for an obtuse angle at C, AC^2 + BC^2 < AB^2 (converse of the Pythagorean theorem).
From CD^2 = AD \cdot DB, \displaystyle \frac{AD}{CD} = \frac{CD}{DB}. The right angles \angle ADC and \angle CDB are congruent and each lies between the two sides in the ratio, so \triangle ADC \sim \triangle CDB by SAS similarity, with A \leftrightarrow C, D \leftrightarrow D, C \leftrightarrow B. So \angle DAC \cong \angle DCB. In right \triangle ADC, m\angle DAC + m\angle ACD = 90^\circ, so m\angle DCB + m\angle ACD = 90^\circ. D lies between A and B, so it is inside \angle ACB and m\angle ACB = m\angle ACD + m\angle DCB = 90^\circ.
P lies on ray BC, so the angle of \triangle PMB at B is \angle ABC, and its angle at M is right, so its angle at P is 90^\circ - m\angle B = m\angle A. Q lies on ray AC, so \triangle AMQ has the angle \angle A at A and a right angle at M. By AA, \triangle AMQ \sim \triangle PMB, with A \leftrightarrow P, M \leftrightarrow M, Q \leftrightarrow B. So \displaystyle \frac{AM}{PM} = \frac{MQ}{MB}, and \displaystyle MP \cdot MQ = AM \cdot MB = \left(\frac{AB}{2}\right)^2. Multiply by 4.
Both have a right angle at D. In right \triangle ADC the acute angles add to 90^\circ: m\angle DAC + m\angle ACD = 90^\circ. D lies inside the right angle at C, so m\angle ACD + m\angle DCB = 90^\circ. Both \angle DAC and \angle DCB are complements of \angle ACD, so they are congruent, and AA gives \triangle ADC \sim \triangle CDB.
From ab = ch, \displaystyle h = \frac{ab}{c}, so \displaystyle \frac{1}{h^2} = \frac{c^2}{a^2 b^2}. Replace c^2 by a^2 + b^2 and split the fraction: \displaystyle \frac{a^2 + b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}. In Example 1, \displaystyle \frac{1}{52} + \frac{1}{117} = \frac{9 + 4}{468} = \frac{1}{36} = \frac{1}{6^2}.
b^2 = cp and a^2 = cq, so \displaystyle \frac{b^2}{a^2} = \frac{cp}{cq} = \frac{p}{q}. In Example 1, \displaystyle \frac{52}{117} = \frac{4}{9}.
Line 2 uses it to make \angle A and \angle B acute, which puts D between A and B; without it the pieces need not add to AB. Lines 4 and 9 use it as the partner of the right angle at D; without it neither AA step has its second pair, and neither similarity follows.