Geometry

Geometric means in a right triangle

Drop a segment from the square corner of a right triangle straight to the long side. How is its length tied to the two pieces it cuts that side into?

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Geometric means in a right triangle

Why this exists

A right triangle with one extra segment drawn in, the altitude from the right angle to the hypotenuse, has six lengths in it: the two legs, the hypotenuse, the altitude, and the two pieces the altitude cuts the hypotenuse into. The Pythagorean theorem ties three of them together, the legs and the hypotenuse, and says nothing about the other three.

The altitude needs a tie of its own. It splits the triangle into two smaller triangles with the same angles as the whole, so the figure holds three similar triangles, and similar triangles give proportions. The proportions that come out have a particular shape: one length appears twice, once at the bottom of a ratio and once at the top of the ratio it equals, as in \displaystyle \frac{p}{h} = \frac{h}{q}. A number that sits in the middle of a proportion like that is called a geometric mean, and this figure is full of them.

The payoff is practical. Any two of the six lengths, with a few limits, give the other four by multiplying and taking square roots, with no angle measured. The altitude fact is old: it appears in Euclid’s Elements as a corollary of Proposition 8 of Book VI, and Proposition 13 of the same book uses it to construct a length that is the “mean proportional” between two given lengths.

Definition

The altitude to the hypotenuse. Let \triangle ABC have its right angle at C, so that \overline{AB} is the hypotenuse. The altitude from C is the segment from C perpendicular to \overline{AB}. It meets \overline{AB} at a point D, the foot of the altitude, and in a right triangle the foot always lands strictly between A and B (the reason is set out here). So the altitude cuts the hypotenuse into two pieces, sometimes called the segments of the hypotenuse. The usual names for the six lengths are

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

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

and since D lies between A and B, the pieces add to the whole: p + q = c.

Figure 1The altitude \overline{CD} cuts the hypotenuse into the pieces p and q, so c = p + q.

Each piece shares an endpoint with one leg. The piece p = AD touches the leg b = AC at A, and the piece q = DB touches the leg a = BC at B. A piece and a leg that share an endpoint are called adjacent.

The geometric mean. For two positive numbers x and y, the geometric mean is the positive number g with

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

Clearing the fractions gives g^2 = xy, so

g = \sqrt{xy}

The older name is the mean proportional, because g is the middle term of the proportion. The geometric mean of 3 and 12 is 6, since \displaystyle \frac{3}{6} = \frac{6}{12}: going from 3 to 6 and from 6 to 12 both multiply by 2. The ordinary average \displaystyle \frac{x + y}{2}, the arithmetic mean, sits halfway between two numbers by adding; the geometric mean sits halfway by multiplying. Both lie between x and y.

Intuition

The altitude splits the right angle at C into two smaller angles, \angle ACD and \angle DCB, which add to 90^\circ. In \triangle ADC the angle at D is right, so \angle A and \angle ACD also add to 90^\circ (the triangle angle sum). Both \angle A and \angle DCB are what \angle ACD needs to make 90^\circ, so they are equal. In the same way \angle ACD equals \angle B. Each small triangle has a right angle, a copy of \angle A and a copy of \angle B: the same shape as the whole triangle, at a smaller size.

Figure 2 lifts the two small triangles out and turns them to face the same way, with the copy of \angle A at the left corner and the right angle on top.

Figure 2The altitude h is the right side of one small triangle and the left side of the other. Similar triangles give \displaystyle \frac{p}{h} = \frac{h}{q}.

The altitude is in both. In \triangle ADC it is the right side; in \triangle CDB it is the left side. Similar triangles have the same ratio of left side to right side, and in the first that ratio is \displaystyle \frac{p}{h}, in the second \displaystyle \frac{h}{q}. So \displaystyle \frac{p}{h} = \frac{h}{q}, and the altitude is the geometric mean of the two pieces. A leg plays the same double part between a small triangle and the whole one, which gives the second rule below.

Two pictures of what h^2 = pq says. In areas: the square with side h has exactly the area of the rectangle with sides p and q. In sizes: when the pieces differ, the altitude is longer than the short piece and shorter than the long one (Exercise 16).

Results

Three similar triangles. The altitude to the hypotenuse cuts a right triangle into two triangles similar to it and to each other. With the letters in matching positions,

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

The altitude rule (geometric mean and the altitude). The altitude to the hypotenuse is the geometric mean of the two pieces of the hypotenuse:

\frac{p}{h} = \frac{h}{q} \qquad h^2 = pq \qquad h = \sqrt{pq}

The leg rule (geometric mean and a leg). Each leg is the geometric mean of the whole hypotenuse and the piece adjacent to that leg:

\frac{c}{b} = \frac{b}{p} \qquad b^2 = cp

\frac{c}{a} = \frac{a}{q} \qquad a^2 = cq

In words: for the altitude, piece times piece; for a leg, whole times its own piece. Together with p + q = c, these let any two of the six lengths find the other four, as long as the two given lengths can belong to a right triangle at all (see Boundaries).

Why it is true

The foot D lies strictly between A and B, so D is on \overline{AB} and inside the right angle at C. Everything else is AA similarity and proportions in similar triangles.

The leg b. \triangle ADC and \triangle ACB each have a right angle, at D and at C, and they share \angle A. By AA, \triangle ADC \sim \triangle ACB, with A \leftrightarrow A, D \leftrightarrow C and C \leftrightarrow B. Corresponding sides are proportional; side \overline{AD} matches \overline{AC} and side \overline{AC} matches \overline{AB}, so

\frac{AD}{AC} = \frac{AC}{AB} \qquad \frac{p}{b} = \frac{b}{c}

and clearing fractions gives b^2 = cp.

The leg a. \triangle CDB and \triangle ACB each have a right angle, at D and at C, and they share \angle B. By AA, \triangle CDB \sim \triangle ACB, with C \leftrightarrow A, D \leftrightarrow C and B \leftrightarrow B. Side \overline{DB} matches \overline{CB} and side \overline{CB} matches \overline{AB}, so \displaystyle \frac{q}{a} = \frac{a}{c} and a^2 = cq.

The altitude. In right \triangle ADC, m\angle A + m\angle ACD = 90^\circ. Ray CD lies inside the right angle at C, so m\angle ACD + m\angle DCB = 90^\circ. Both \angle A and \angle DCB are complements of \angle ACD, so \angle A \cong \angle DCB. The two small triangles also have right angles at D, so AA gives \triangle ADC \sim \triangle CDB with A \leftrightarrow C, D \leftrightarrow D and C \leftrightarrow B. Side \overline{AD} matches \overline{CD} and side \overline{DC} matches \overline{DB}, so

\frac{AD}{CD} = \frac{DC}{DB} \qquad \frac{p}{h} = \frac{h}{q}

and h^2 = pq.

Adding the two leg rules gives a^2 + b^2 = cq + cp = c(p + q) = c^2, which is a whole proof of the Pythagorean theorem; it is written out with every reason in proving the Pythagorean theorem with similar triangles.

Worked examples

Example 1

The geometric mean of two numbers

Find the geometric mean of (a) 3 and 27, (b) 6 and 10.

Solution

(a) The geometric mean g satisfies g^2 = 3 \cdot 27 = 81, and g is positive, so g = 9. Check it as a proportion: \displaystyle \frac{3}{9} = \frac{9}{27}, since both equal \displaystyle \frac{1}{3}. Each step from 3 to 9 to 27 multiplies by 3. The ordinary average of 3 and 27 is 15, a different number: halfway by adding is not halfway by multiplying.

(b) Here g^2 = 6 \cdot 10 = 60, so \displaystyle g = \sqrt{60}. Since 60 = 4 \cdot 15 and 4 is a perfect square, the square root simplifies:

g = \sqrt{60} = 2\sqrt{15} \approx 7.75

It lies between 6 and 10, a little below their average 8.

Example 2

Three lengths from the two pieces

In Figure 3, \triangle ABC has its right angle at C, and the altitude \overline{CD} cuts the hypotenuse into AD = 4 and DB = 12. Find x, y and z.

Figure 3The altitude cuts the hypotenuse into 4 and 12. Find x, y and z.

Solution

x is the altitude, so it is the geometric mean of the two pieces:

\begin{aligned} x^2 &= 4 \cdot 12 = 48 \\ x &= \sqrt{48} = 4\sqrt{3} \end{aligned}

y and z are legs, so each needs the whole hypotenuse, 4 + 12 = 16, and its own adjacent piece. y = AC touches the piece AD = 4 at A, and z = BC touches the piece DB = 12 at B:

\begin{aligned} y^2 &= 16 \cdot 4 = 64, &\quad y &= 8 \\ z^2 &= 16 \cdot 12 = 192, &\quad z &= 8\sqrt{3} \end{aligned}

Check. y^2 + z^2 = 64 + 192 = 256 = 16^2, as the Pythagorean theorem requires. The leg y = 8 is half the hypotenuse, so this is the 30^\circ-60^\circ-90^\circ triangle, one of the special right triangles, with sides in the ratio \displaystyle 1 : \sqrt{3} : 2.

Example 3

From a leg and its piece

In right \triangle ABC with the right angle at C and altitude \overline{CD}, the leg AC = 6 and the piece AD = 4. Find AB, DB, CD and BC.

Solution

The leg rule runs in either direction. AC is the geometric mean of the hypotenuse and its adjacent piece AD, so

\begin{aligned} AC^2 &= AB \cdot AD \\ 36 &= AB \cdot 4 \\ AB &= 9 \end{aligned}

The pieces add to the hypotenuse, so DB = 9 - 4 = 5. Now both pieces are known, and the other two lengths follow:

\begin{aligned} CD^2 &= AD \cdot DB = 4 \cdot 5 = 20, &\quad CD &= 2\sqrt{5} \\ BC^2 &= AB \cdot DB = 9 \cdot 5 = 45, &\quad BC &= 3\sqrt{5} \end{aligned}

Check. AC^2 + BC^2 = 36 + 45 = 81 = 9^2.

Example 4

A piece that is a fraction

In right \triangle ABC with the right angle at C, the altitude CD = 4 and the piece AD = 6. Find DB, AB, AC and BC.

Solution

The altitude rule, with one piece known, gives the other:

\begin{aligned} CD^2 &= AD \cdot DB \\ 16 &= 6 \cdot DB \\ DB &= \frac{16}{6} = \frac{8}{3} \end{aligned}

Nothing requires a length to be a whole number. The hypotenuse is \displaystyle AB = 6 + \frac{8}{3} = \frac{26}{3}, and the leg rule gives each leg:

\begin{aligned} AC^2 &= \tfrac{26}{3} \cdot 6 = 52, &\quad AC &= 2\sqrt{13} \\ BC^2 &= \tfrac{26}{3} \cdot \tfrac{8}{3} = \tfrac{208}{9}, &\quad BC &= \tfrac{4\sqrt{13}}{3} \end{aligned}

For BC, \displaystyle \sqrt{208} = \sqrt{16 \cdot 13} = 4\sqrt{13} and \displaystyle \sqrt{9} = 3.

Check. \displaystyle AC^2 + BC^2 = \frac{468}{9} + \frac{208}{9} = \frac{676}{9}, and \displaystyle AB^2 = \left(\frac{26}{3}\right)^2 = \frac{676}{9}. The altitude 4 also lies between the pieces \displaystyle \frac{8}{3} and 6, as it must.

Example 5

A quadratic from the leg rule

In right \triangle ABC with the right angle at C and altitude \overline{CD}, the leg AC = 15 and the piece DB = 16. Find AD, AB, BC and CD.

Solution

The given leg is AC, but the given piece DB is not adjacent to it, so neither rule can be used straight away. Name the unknown: let AD = x, so AB = x + 16. The leg rule for AC now becomes an equation in x:

\begin{aligned} x(x + 16) &= 225 \\ x^2 + 16x - 225 &= 0 \\ (x + 25)(x - 9) &= 0 \end{aligned}

By the zero product property, x = 9 or x = -25. A length is positive, so AD = 9. Then AB = 25, and

\begin{aligned} BC^2 &= AB \cdot DB = 25 \cdot 16 = 400, &\quad BC &= 20 \\ CD^2 &= AD \cdot DB = 9 \cdot 16 = 144, &\quad CD &= 12 \end{aligned}

Check. 15^2 + 20^2 = 225 + 400 = 625 = 25^2. Twice the area two ways also agrees: the legs give 15 \cdot 20 = 300, and the hypotenuse with the altitude gives 25 \cdot 12 = 300.

Example 6

A roof truss

Two rafters meet at a right angle at the ridge of a roof, and their lower ends rest on the two ends of a horizontal beam. A vertical post runs from the ridge down to the beam and meets it 2.5 m from one end and 10 m from the other. How tall is the post, and how long is each rafter?

Solution

First match the roof to the figure. The rafters and the beam form a triangle with its right angle at the ridge, so the rafters are the legs and the beam is the hypotenuse. The post starts at the right angle, and it is vertical while the beam is horizontal, so it is perpendicular to the beam: it is the altitude to the hypotenuse, and its foot cuts the beam into pieces of 2.5 m and 10 m.

The altitude rule gives the post:

h^2 = 2.5 \cdot 10 = 25 \qquad\Longrightarrow\qquad h = 5

The beam is 2.5 + 10 = 12.5 m long, and each rafter is the geometric mean of the beam and the piece at its own end:

\begin{aligned} \text{short rafter}^2 &= 12.5 \cdot 2.5 = 31.25 \\ \text{long rafter}^2 &= 12.5 \cdot 10 = 125 \end{aligned}

So the post is 5 m tall and the rafters are \displaystyle \sqrt{31.25} = \frac{5\sqrt{5}}{2} \approx 5.59 m and \displaystyle \sqrt{125} = 5\sqrt{5} \approx 11.18 m.

Check. 31.25 + 125 = 156.25 = 12.5^2.

Example 7

The radius of an arch

A bridge arch is part of a circle. It spans 24 m and rises 4 m above the span at its middle. Find the radius of the circle.

Solution

Figure 4 names the points. The span is the chord \overline{EF} with midpoint M, and T is the top of the arch, 4 m straight above M. The perpendicular from the center of the circle to the chord cuts the chord in half (chord theorems), so it meets \overline{EF} at its midpoint M. Only one line is perpendicular to \overline{EF} at M, so the vertical line through M and T passes through the center. It therefore holds a whole diameter: it meets the circle at T and again at a point S below, and \overline{TS} is a diameter.

Figure 4The arch is part of a circle. The right triangle TES has \overline{EM} as the altitude to its hypotenuse.

The arch hides a right triangle. E lies on the circle and \overline{TS} is a diameter, so \angle TES is a right angle (an angle in a half circle is right). In right \triangle TES, the segment \overline{EM} runs from the right angle at E and is perpendicular to the hypotenuse \overline{TS}, so it is the altitude to the hypotenuse, and M cuts the hypotenuse into TM = 4 and MS. The altitude is half the span, EM = 12.

The altitude rule:

\begin{aligned} EM^2 &= TM \cdot MS \\ 144 &= 4 \cdot MS \\ MS &= 36 \end{aligned}

The diameter is TS = 4 + 36 = 40, so the radius is 20 m.

Check. The center O is 20 m below T, so OM = 20 - 4 = 16, and OE^2 = OM^2 + EM^2 = 16^2 + 12^2 = 400 = 20^2. The end of the span is 20 m from the center, so it does lie on the circle.

Boundaries

The angle at C must be right. In any triangle the altitude from C meets line AB at a right angle, but the three similar triangles need the right angle at C as the partner of the right angle at D. With D between A and B, h^2 = pq holds exactly when the angle at C is right (the converse is Exercise 17 of the proof lesson). Raise C further from \overline{AB} and the angle at C narrows while h^2 grows past pq; lower it and the angle widens while h^2 falls below pq. Exercise 17 below is an example.

Each leg goes with its own piece. The leg rule pairs a leg with the piece that shares its endpoint, because those two are corresponding sides of the same pair of similar triangles. The other piece is not a side of the small triangle that contains that leg, and the product c times the other piece is the square of the other leg.

Only the altitude from the right angle. A right triangle has three altitudes, but the altitude from A to line BC is the leg \overline{AC} itself, and the same for B. Neither cuts its side into two pieces, and there is nothing new to say about them (Exercise 21).

Not every pair of lengths fits. A leg is shorter than the hypotenuse, and the piece adjacent to a leg is shorter than that leg, since \displaystyle p = \frac{b^2}{c} and \displaystyle \frac{b}{c} < 1. The altitude can be at most half the hypotenuse, because the geometric mean of two numbers is never more than their average (shown with a half circle there). Given lengths that break one of these limits describe no right triangle, and the algebra says so (Exercise 22).

A geometric mean is taken of positive numbers. The equation g^2 = xy has a negative solution too, \displaystyle -\sqrt{xy}, but a length is positive, so the geometric mean is the positive root. When an unknown piece comes from a quadratic, as in Example 5, a negative root is discarded for the same reason.

Where this goes next

The two leg rules, added, are a proof of the Pythagorean theorem that uses no area at all.

The leg rule written as \displaystyle \frac{p}{b} = \frac{b}{c} says that two ratios inside two right triangles sharing \angle A are the same number. That number depends only on \angle A, and it is the cosine of \angle A (right triangle trigonometry). Similar right triangles having equal ratios of sides is the whole foundation of trigonometry.

Example 7 is a circle theorem in disguise. The chords \overline{EF} and \overline{TS} cross at M, and the rule for two crossing chords says EM \cdot MF = TM \cdot MS (segment lengths in circles, proved in proving the power of a point relationships). When one chord is cut in half by a diameter, EM = MF and the chord rule becomes the altitude rule.

Exercises

Geometric means

Find the geometric mean of each pair of numbers. Give exact answers.

4 and 25
2 and 32
5 and 15
\displaystyle \frac{1}{2} and 18
The geometric mean of 6 and a positive number n is 12. Find n.

Finding lengths

In Exercises 6 to 13, \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 = 2 and DB = 8. Find CD.
AD = 3 and DB = 15. Find CD, AC and BC.
AD = 1.8 and AB = 5. Find AC, DB, CD and BC.
CD = 6 and AD = 3. Find DB, AB, AC and BC.
BC = 10 and DB = 8. Find AB, AD, CD and AC.
AC = 5 and BC = 12. Find AB, AD, DB and CD.
CD = 8 and AB = 20. Find AD, DB, AC and BC.
AD = x, DB = 2x and CD = 4. Find x, AB, AC and BC.

In context

A field is a right triangle. A straight path runs from the corner with the right angle to the longest side, meeting it at a right angle, 48 m from one end of that side and 27 m from the other. How long is the path, and how long are the two shorter sides of the field?
A circular arch spans 16 m and rises 4 m above the span at its middle. Find the radius of the circle.

Reasoning

Show that when the two pieces of the hypotenuse are different lengths, the altitude is longer than the shorter piece and shorter than the longer one.
In \triangle ABC, \overline{CD} \perp \overline{AB} with D between A and B, AD = DB = 2 and CD = 3. Is \angle ACB a right angle? If not, is it acute or obtuse? Explain.
Use the similarity \triangle CDB \sim \triangle ACB to explain why the leg \overline{BC} pairs with the piece \overline{DB} in the leg rule, and not with \overline{AD}.
Multiply the two leg rules together and use the altitude rule to show that a^2 b^2 = c^2 h^2, and so ab = ch. What do the two sides of ab = ch measure?
The altitude to the hypotenuse cuts it into two equal pieces. Show that the altitude equals each piece and that the triangle is isosceles. What are its angles?
In right \triangle ABC with the right angle at C, what is the altitude from A to line BC? Explain why it gives no geometric mean relation.
Can a right triangle have an altitude of 7 to a hypotenuse of 12? Explain.

Answers

10. g^2 = 4 \cdot 25 = 100.
8. g^2 = 2 \cdot 32 = 64.
\displaystyle 5\sqrt{3}. g^2 = 5 \cdot 15 = 75 = 25 \cdot 3.
3. \displaystyle g^2 = \frac{1}{2} \cdot 18 = 9.
n = 24. The proportion is \displaystyle \frac{6}{12} = \frac{12}{n}, so 6n = 144.
CD = 4. CD^2 = 2 \cdot 8 = 16.
\displaystyle CD = 3\sqrt{5}, \displaystyle AC = 3\sqrt{6}, \displaystyle BC = 3\sqrt{30}. AB = 18; CD^2 = 3 \cdot 15 = 45; AC^2 = 18 \cdot 3 = 54; BC^2 = 18 \cdot 15 = 270. Check: 54 + 270 = 324 = 18^2.
AC = 3, DB = 3.2, CD = 2.4, BC = 4. AC^2 = 5 \cdot 1.8 = 9; DB = 5 - 1.8; CD^2 = 1.8 \cdot 3.2 = 5.76; BC^2 = 5 \cdot 3.2 = 16. The triangle has sides 3, 4, 5.
DB = 12, AB = 15, \displaystyle AC = 3\sqrt{5}, \displaystyle BC = 6\sqrt{5}. \displaystyle DB = \frac{CD^2}{AD} = \frac{36}{3}; AC^2 = 15 \cdot 3 = 45; BC^2 = 15 \cdot 12 = 180. Check: 45 + 180 = 225 = 15^2.
AB = 12.5, AD = 4.5, CD = 6, AC = 7.5. \displaystyle AB = \frac{BC^2}{DB} = \frac{100}{8}; AD = 12.5 - 8; CD^2 = 4.5 \cdot 8 = 36; AC^2 = 12.5 \cdot 4.5 = 56.25. Check: 56.25 + 100 = 156.25 = 12.5^2.
AB = 13, \displaystyle AD = \frac{25}{13}, \displaystyle DB = \frac{144}{13}, \displaystyle CD = \frac{60}{13}. AB^2 = 25 + 144 = 169; \displaystyle AD = \frac{AC^2}{AB} and \displaystyle DB = \frac{BC^2}{AB}; \displaystyle CD^2 = AD \cdot DB = \frac{3600}{169}. The pieces add to \displaystyle \frac{169}{13} = 13.
AD = 4 and DB = 16 (or the other way round, which is the same triangle turned over), \displaystyle AC = 4\sqrt{5}, \displaystyle BC = 8\sqrt{5}. With AD = x: x(20 - x) = 64, so x^2 - 20x + 64 = 0 and (x - 4)(x - 16) = 0. Then AC^2 = 20 \cdot 4 = 80 and BC^2 = 20 \cdot 16 = 320. Check: 80 + 320 = 400 = 20^2.
\displaystyle x = 2\sqrt{2}, \displaystyle AB = 6\sqrt{2}, \displaystyle AC = 2\sqrt{6}, \displaystyle BC = 4\sqrt{3}. The altitude rule gives 16 = x \cdot 2x = 2x^2, so x^2 = 8 and, as a length is positive, \displaystyle x = 2\sqrt{2}. Then AB = 3x, \displaystyle AC^2 = 6\sqrt{2} \cdot 2\sqrt{2} = 24 and \displaystyle BC^2 = 6\sqrt{2} \cdot 4\sqrt{2} = 48. Check: \displaystyle 24 + 48 = 72 = (6\sqrt{2})^2.
The path is 36 m; the shorter sides are 45 m and 60 m. The path starts at the right angle and is perpendicular to the longest side, so it is the altitude to the hypotenuse: h^2 = 48 \cdot 27 = 1296. The longest side is 75 m, so the side ending at the 27 m piece has square 75 \cdot 27 = 2025, and the other 75 \cdot 48 = 3600.
10 m. As in Example 7, half the span is the altitude of a right triangle inscribed in the circle: 8^2 = 4 \cdot MS, so MS = 16, the diameter is 4 + 16 = 20 and the radius 10.
Call the pieces p < q. Multiplying p < q by the positive number p gives p^2 < pq, and multiplying it by q gives pq < q^2. Since h^2 = pq, this says p^2 < h^2 < q^2, and for positive numbers squares keep their order, so p < h < q.
No; it is acute. If it were right, the altitude rule would give CD^2 = AD \cdot DB = 4, but CD^2 = 9. Each of \triangle ADC and \triangle CDB is a right triangle with legs 2 and 3, and its angle at C faces the shorter leg, so it is the smaller of two acute angles that add to 90^\circ (sides and angles in order). Each is less than 45^\circ, so \angle ACB is less than 90^\circ. This fits the rule in Boundaries: here h^2 = 9 > 4 = pq.
The correspondence is C \leftrightarrow A, D \leftrightarrow C, B \leftrightarrow B, so \overline{DB} matches \overline{CB} and \overline{CB} matches \overline{AB}, which gives \displaystyle \frac{DB}{CB} = \frac{CB}{AB} and BC^2 = AB \cdot DB. The small triangle containing the leg \overline{BC} is \triangle CDB, and the only piece of the hypotenuse inside it is \overline{DB}, the one touching B. \overline{AD} is not a side of that triangle.
a^2 b^2 = (cq)(cp) = c^2 (pq) = c^2 h^2. All four lengths are positive, so taking square roots gives ab = ch. Each side is twice the area of \triangle ABC: once with the legs as base and height, once with the hypotenuse as base and the altitude as height.
With p = q, the altitude rule gives h^2 = p^2, so h = p. The leg rule gives a^2 = cq = cp = b^2, so a = b and the triangle is isosceles; its angles are 45^\circ, 45^\circ and 90^\circ (special right triangles). Then D is the midpoint of \overline{AB}, and DA = DB = DC.
It is the leg \overline{AC}. \overline{AC} \perp \overline{BC} already, so the perpendicular from A to line BC lands at C, the end of \overline{BC}. It does not cut \overline{BC} into two pieces, and the triangle it would cut off has no area, so there is no pair of similar triangles to compare.
No. If the pieces are x and 12 - x, the altitude rule needs x(12 - x) = 49, that is x^2 - 12x + 49 = 0, and the discriminant is 144 - 196 = -52, which is negative, so no real x works. Another way to see it: the altitude can be at most half the hypotenuse, which is 6.