Geometry · G.GPE.B.4

Coordinate proofs with variables

Checking a theorem on one triangle with numbers proves nothing about the next triangle. How do you put every triangle on the grid at once, and where should it go so the algebra stays short?

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The midpoint formulaGeometry The distance formulaMath 8 Midsegments of triangles and trapezoidsGeometry

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Coordinate proofs with variables

When you reach for it

The midsegment theorem says that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. You can test it on the triangle with vertices (2, 8), (-4, 0) and (10, 2): find the midpoints, compare slopes, compare lengths, and it works. But that is one triangle. There are infinitely many others, and no number of numerical checks covers them all.

A coordinate proof fixes this by putting a triangle on the grid with letters for its coordinates instead of numbers. The letters can stand for any values, so a computation done with them is done for every triangle at once. The distance formula, the midpoint formula and the slope formula then turn statements about the figure into algebra:

two segments are congruent when the distance formula gives the same length;
two segments bisect each other when they have the same midpoint;
two lines are parallel when their slopes are equal, and perpendicular when the product of their slopes is -1 (parallel lines, perpendicular lines).

Reach for a coordinate proof when the theorem is about lengths, midpoints, parallel lines or right angles. The skill this lesson is about is the step before the algebra: where to put the figure. A good placement leaves the figure completely general and still makes most coordinates zero. A bad one either drowns the proof in letters or, worse, quietly proves the theorem for only some figures.

The procedure

Why the figure may be moved

Every coordinate proof starts by deciding where the figure sits, and the theorem does not say where. That freedom is real. Sliding, turning and flipping a figure are rigid motions, and they change no length and no angle. So they keep midpoints as midpoints, parallel lines parallel and perpendicular lines perpendicular. A theorem about those things is true of a figure exactly when it is true of the moved figure.

So any triangle ABC can be slid until A is at the origin, turned until B lies on the positive x-axis, and, if C then lies below the axis, flipped across the x-axis so that C is above it. Nothing about the triangle has been lost. That is why a proof may simply write A(0, 0) and B(a, 0) without first saying “suppose the triangle happens to sit there”.

Put the origin and the axes where the figure is simplest

Zeros make the algebra short, so choose the origin and the axes to create as many as possible.

Put the origin at a vertex and run one side along the positive x-axis. That vertex becomes (0, 0) and the next one (a, 0).
At a right angle, put the vertex at the origin and the two sides along both axes. Every vertex on a side then has a zero in it.
When the figure is symmetric, as an isosceles triangle is, put the line of symmetry on the y-axis. Points then come in pairs (x, y) and (-x, y), and many sums cancel.

Give every free coordinate a letter, and no more

A placement is general when every figure that fits the Given can be moved to sit that way. The test is to count. A triangle is pinned down by its three side lengths, so a general triangle needs three free letters: A(0, 0), B(a, 0), C(b, c) with a > 0, c > 0 and b any number. (b may be negative: then the angle at A is obtuse.) A right triangle is pinned down by its two legs, so it needs only two.

Figure	A general placement	Free letters
any triangle	(0, 0), (a, 0), (b, c)	3
right triangle	(0, 0), (a, 0), (0, b)	2
isosceles triangle	(-a, 0), (a, 0), (0, b)	2
equilateral triangle	(-a, 0), (a, 0), \displaystyle (0, a\sqrt{3})	1
parallelogram	(0, 0), (a, 0), (a + b, c), (b, c)	3
rhombus	the parallelogram, with b^2 + c^2 = a^2	2
rectangle	(0, 0), (a, 0), (a, b), (0, b)	2
square	(0, 0), (a, 0), (a, a), (0, a)	1

In every row the letters stand for positive numbers, except the b in the triangle, parallelogram and rhombus rows, which may be any number. Too many letters is safe but slow. Too few is a mistake: the proof then covers only the figures that happen to fit, such as right triangles when the theorem was about all triangles.

Let the Given fix the rest

A coordinate that the definition forces is worked out, not chosen. In a parallelogram with A(0, 0), B(a, 0) and D(b, c), the fourth vertex is not free: opposite sides must be parallel, and that forces C(a + b, c), as Example 3 shows. A condition such as “the sides are equal” becomes an equation between the letters, which is how the rhombus row above gets its b^2 + c^2 = a^2. Write every such condition down, with every inequality (a > 0, c > 0), because the proof may need it later to rule out a zero denominator.

Use factors of 2 when midpoints are coming

The midpoint formula halves sums. With B(a, 0) and C(b, c) the midpoint of \overline{BC} is \displaystyle \left(\frac{a + b}{2}, \frac{c}{2}\right), and those halves ride through every later step. Writing the vertices as B(2a, 0) and C(2b, 2c) instead gives the midpoint (a + b, c). Nothing is lost, because every number is twice some number: 2a can still be any positive value. If a proof will need thirds, as a centroid does, use factors of 6, which make both halves and thirds clean.

Compute, then say what the computation means

With the figure placed, each claim in the Prove becomes one calculation: lengths by the distance formula, a shared point by the midpoint formula, parallel or perpendicular by slopes. Before dividing by a run, make sure it cannot be 0, or treat the vertical case separately. The last sentence translates back: not “\displaystyle \sqrt{a^2 + b^2} = \sqrt{a^2 + b^2}” but “so MP = MR”.

A numerical case is still useful, just not as a proof. Choosing values for the letters is a quick way to test a claim before proving it, and to check the algebra after.

Worked examples

Example 1

Choosing coordinates for three triangles

Give a general placement, with as few letters as possible, for (a) a triangle PQR with a right angle at R; (b) a triangle ABC with AB = AC; (c) any triangle ABC whose midpoints will be needed.

Solution

(a) Put the right angle at the origin. Turn the triangle so \overline{RP} lies along the positive x-axis. The angle at R is a right angle, so \overline{RQ} lies along the y-axis, and a flip across the x-axis, if needed, puts Q above. That gives

R(0, 0), \qquad P(p, 0), \qquad Q(0, q)

with p > 0 and q > 0. Two letters, the two legs, and two numbers are exactly what it takes to pin down a right triangle.

(b) This triangle is symmetric, so put \overline{BC} on the x-axis with the origin at its midpoint: B(-a, 0) and C(a, 0) with a > 0. Do not guess where A goes. Call it A(x, y), with y > 0, and use the Given, AB = AC, squared on both sides so that the square roots go:

\begin{aligned} (x + a)^2 + y^2 &= (x - a)^2 + y^2 \\ x^2 + 2ax + a^2 &= x^2 - 2ax + a^2 \\ 4ax &= 0 \end{aligned}

Since a > 0, this forces x = 0. So A lies on the y-axis, A(0, h) with h > 0, and the placement is B(-a, 0), C(a, 0), A(0, h): two letters, half the base and the height.

A(0, 0), \qquad B(2a, 0), \qquad C(2b, 2c)

with a > 0 and c > 0. Three letters, matching the three side lengths that pin down a triangle. b is allowed to be any number, negative included, since C may sit to the left of A.

Example 2

The midpoint of the hypotenuse

Prove that the midpoint of the hypotenuse of a right triangle is the same distance from all three vertices.

Figure 1The right angle at the origin and the legs on the axes. Two letters describe every right triangle.

Solution

Given. \triangle PQR with a right angle at R; M is the midpoint of \overline{PQ}.

Prove. MP = MQ = MR.

Place the triangle as in Example 1(a), with factors of 2 because a midpoint is coming: R(0, 0), P(2a, 0), Q(0, 2b), a > 0, b > 0. By the midpoint formula,

M = \left(\frac{2a + 0}{2},\ \frac{0 + 2b}{2}\right) = (a, b)

Now the three distances from M, each by the distance formula:

\begin{aligned} MR &= \sqrt{(a - 0)^2 + (b - 0)^2} = \sqrt{a^2 + b^2} \\ MP &= \sqrt{(2a - a)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} \\ MQ &= \sqrt{(0 - a)^2 + (2b - b)^2} = \sqrt{a^2 + b^2} \end{aligned}

All three are \displaystyle \sqrt{a^2 + b^2}, so MP = MQ = MR. Since a and b stand for any two positive numbers, this covers every right triangle.

Check with numbers: legs 6 and 8 mean a = 3 and b = 4, so each distance is \displaystyle \sqrt{9 + 16} = 5, and indeed the hypotenuse is 10 and M is halfway along it. The theorem says a circle centered at M passes through all three vertices, so the circumcenter of a right triangle is the midpoint of its hypotenuse.

Example 3

The diagonals of a parallelogram

Prove that the diagonals of a parallelogram bisect each other.

Figure 2Three letters describe every parallelogram. The fourth vertex is forced by the other three.

Solution

Given. ABCD is a parallelogram.

Prove. \overline{AC} and \overline{BD} have the same midpoint.

Placing it. Put A at the origin and B on the positive x-axis, and flip if needed so that D is above the axis. Midpoints are coming, so use factors of 2: A(0, 0), B(2a, 0), D(2b, 2c), with a > 0, c > 0 and b any number. Three letters, as a parallelogram needs.

C is not free, so work it out from the definition: both pairs of opposite sides are parallel. Call it C(x, y).

\overline{DC} \parallel \overline{AB}, and \overline{AB} is horizontal, so \overline{DC} is horizontal too. C is at the same height as D: y = 2c.
\overline{BC} \parallel \overline{AD}. If b \ne 0, the slopes are equal: \displaystyle \frac{2c - 0}{x - 2a} = \frac{2c}{2b}, so x - 2a = 2b and x = 2a + 2b. If b = 0, \overline{AD} is vertical, so \overline{BC} is vertical and x = 2a, which is again 2a + 2b.

So C(2a + 2b, 2c) in every case.

The proof. Call the midpoints of the diagonals M_{AC} and M_{BD}. By the midpoint formula,

\begin{aligned} M_{AC} &= \left(\frac{0 + 2a + 2b}{2},\ \frac{0 + 2c}{2}\right) = (a + b,\ c) \\ M_{BD} &= \left(\frac{2a + 2b}{2},\ \frac{0 + 2c}{2}\right) = (a + b,\ c) \end{aligned}

The two diagonals have the same midpoint, M(a + b, c), so each one cuts the other into two equal halves: the diagonals bisect each other, one of the properties of a parallelogram, now proved for every parallelogram at once.

Example 4

The medians of an isosceles triangle

In \triangle ABC, AB = AC. Prove that the medians to the two equal sides are congruent.

Figure 3An isosceles triangle placed symmetrically about the y-axis. The two medians to the equal sides have the same length.

Solution

Given. AB = AC; M is the midpoint of \overline{AB} and N the midpoint of \overline{AC}.

Prove. CM = BN.

Example 1(b) showed that the apex of an isosceles triangle placed this way lands on the y-axis. Midpoints are coming, so double the letters: B(-2a, 0), C(2a, 0), A(0, 2b) with a > 0, b > 0. Then

\begin{aligned} M &= \left(\frac{-2a + 0}{2},\ \frac{0 + 2b}{2}\right) = (-a,\ b) \\ N &= \left(\frac{2a + 0}{2},\ \frac{0 + 2b}{2}\right) = (a,\ b) \end{aligned}

The median from C ends at M, and the median from B ends at N:

\begin{aligned} CM &= \sqrt{(-a - 2a)^2 + (b - 0)^2} \\ &= \sqrt{9a^2 + b^2} \\ BN &= \sqrt{(a - (-2a))^2 + (b - 0)^2} \\ &= \sqrt{9a^2 + b^2} \end{aligned}

So CM = BN. The symmetric placement did most of the work: M and N are mirror images, so the two computations differ only in signs that the squaring removes. With A at the origin instead, the same proof needs an extra equation to express AB = AC, and every distance carries more terms.

Example 5

The midsegment theorem with no placement at all

Prove the midsegment theorem using a completely general triangle, with no vertex placed anywhere special. Then compare the work with a good placement.

Solution

Given. \triangle ABC; D is the midpoint of \overline{AB} and E the midpoint of \overline{AC}.

Prove. \overline{DE} \parallel \overline{BC} and \displaystyle DE = \frac{1}{2}BC.

Give each vertex two letters, doubled for the midpoints: A(2p, 2q), B(2r, 2s), C(2t, 2u). Six letters is more than the three a triangle needs, but more letters are never wrong, only slower. The midpoints are

D = (p + r,\ q + s), \qquad E = (p + t,\ q + u)

Length. From D to E the run is t - r and the rise is u - s. From B to C the run is 2t - 2r = 2(t - r) and the rise is 2(u - s). So

\begin{aligned} DE &= \sqrt{(t - r)^2 + (u - s)^2} \\ BC &= \sqrt{4(t - r)^2 + 4(u - s)^2} \\ &= 2\sqrt{(t - r)^2 + (u - s)^2} \end{aligned}

and \displaystyle DE = \frac{1}{2}BC.

Direction. The slope needs a non-zero run, so there are two cases. If t \ne r, both slopes are \displaystyle \frac{u - s}{t - r}, since the factor 2 cancels in \displaystyle \frac{2(u - s)}{2(t - r)}. If t = r, both runs are 0 and both segments are vertical. Either way the two lines have the same direction. They are not one and the same line, because then D would lie on line BC, and the line through B and D would contain A as well, so the triangle would be flat. So \overline{DE} \parallel \overline{BC}.

Comparison. Put B at the origin and C on the positive x-axis, B(0, 0), C(2a, 0), A(2b, 2c), and the vertical case never arises: \overline{BC} is horizontal, D(b, c) and E(a + b, c) are at the same height, and both lengths come straight from subtracting x-coordinates. That shorter proof is written out in two columns in choosing a proof format. Both proofs are correct. The placement only saves work.

Example 6

The diagonals of a rhombus

Prove that the diagonals of a rhombus are perpendicular.

Figure 4A rhombus is a parallelogram with one more condition, b^2 + c^2 = a^2. That condition is what makes the diagonals perpendicular.

Solution

Given. ABCD is a rhombus: a parallelogram with all four sides equal.

Prove. \overline{AC} \perp \overline{BD}.

A rhombus is a parallelogram, so start from the parallelogram placement of Example 3, without the factors of 2 since no midpoint is needed: A(0, 0), B(a, 0), D(b, c), C(a + b, c), with a > 0, c > 0. The extra condition AD = AB becomes an equation:

\sqrt{b^2 + c^2} = a \qquad\Longrightarrow\qquad b^2 + c^2 = a^2

The slopes of the diagonals are

\begin{aligned} \text{slope of } \overline{AC} &= \frac{c}{a + b} \\ \text{slope of } \overline{BD} &= \frac{c - 0}{b - a} = \frac{c}{b - a} \end{aligned}

These need a + b \ne 0 and b \ne a. Both hold: if b = a or b = -a, then b^2 = a^2, and b^2 + c^2 = a^2 would give c = 0, but c > 0. Now multiply, and use the condition in the form b^2 - a^2 = -c^2:

\frac{c}{a + b} \cdot \frac{c}{b - a} = \frac{c^2}{b^2 - a^2} = \frac{c^2}{-c^2} = -1

The product of the slopes is -1, so the diagonals are perpendicular. The equation b^2 + c^2 = a^2 was the whole proof. Without it, which is to say for a parallelogram that is not a rhombus, the product is not -1 (Exercise 13).

Example 7

Which parallelograms have congruent diagonals?

A rectangle’s diagonals are congruent. Use coordinates to find every parallelogram whose diagonals are congruent.

Solution

A coordinate placement can do more than confirm a theorem: it can say exactly when a statement is true. Place a general parallelogram as in Example 6, A(0, 0), B(a, 0), D(b, c), C(a + b, c), with a > 0, c > 0 and b any number, but no condition linking them. Compare the squared lengths, which is simpler and equivalent, since lengths are positive:

\begin{aligned} AC^2 &= (a + b)^2 + c^2 \\ &= a^2 + 2ab + b^2 + c^2 \\ BD^2 &= (b - a)^2 + c^2 \\ &= a^2 - 2ab + b^2 + c^2 \end{aligned}

They differ by AC^2 - BD^2 = 4ab. Since a > 0, this is 0 exactly when b = 0. Then D = (0, c), so \overline{AD} lies along the y-axis, at right angles to \overline{AB}, and the vertices are (0, 0), (a, 0), (a, c), (0, c): a rectangle.

So a parallelogram has congruent diagonals exactly when it is a rectangle. One computation gave both the theorem about rectangles and its converse, the fact behind proving a parallelogram is a rectangle. It also says which diagonal is longer: when b > 0 the angle at A is acute and \overline{AC} is the longer one.

Boundaries

The placement must be general. Every figure that fits the Given has to be movable onto it by sliding, turning and flipping. A right triangle placed with its legs on the axes is general for right triangles and special for triangles; a proof built on it says nothing about obtuse triangles. Counting the free letters against the numbers that pin the figure down is the quickest test.

Only conditions the Given forces may be imposed. a > 0 and c > 0 come from the moves themselves (a side along the positive axis, a vertex flipped above it). b > 0 in a general triangle does not: it would leave out every triangle with an obtuse angle at A.

A slope needs a non-zero run. Before using slopes, show the run cannot be 0, as in Example 6, or treat vertical lines as a separate case, as in Example 5. A placement that puts a side on an axis often makes the question disappear.

Coordinates suit lengths, midpoints, parallel and perpendicular lines. Those have simple formulas. The size of an angle other than a right angle does not, so a theorem such as the angle sum of a triangle is proved without coordinates.

Numbers check, letters prove. A placement with numbers, however carefully chosen, is one figure. It can suggest a theorem or catch an algebra slip, never prove one.

Where this goes next

A coordinate proof is one of the four ways to set out a proof compared in choosing a proof format, which writes the midsegment theorem in two columns and as a flowchart from the placement used here. The parallelogram placement of Examples 3 and 6 is the working tool for proofs about parallelograms and for proving that a parallelogram is a rectangle, and classifying a quadrilateral from coordinates runs the same tests on numbers instead of letters.

Factors of 6 make thirds clean, which is how the centroid of a triangle is shown to lie two thirds of the way along every median (Exercise 11), using the partition formula. Further on, the same habit of choosing axes to match a figure sets a circle’s center at the origin in the equation of a circle, and it runs through all of analytic geometry, where curves as well as polygons are studied through the equations of their points.

Exercises

Placing a figure

A square PQRS has side s. Place it with P at the origin, Q on the positive x-axis and S above it. Give the coordinates of all four vertices and the length of the diagonal \overline{PR}.
A triangle XYZ has XY = XZ, base \overline{YZ} of length 2a and height h. Place \overline{YZ} on the x-axis with the origin at its midpoint and X above. Give the coordinates and the length of each equal side.
An equilateral triangle ABC has A(0, 0), B(2a, 0), a > 0, and C above the x-axis. Find the coordinates of C.
ABCD is a parallelogram with A(0, 0), B(7, 0) and D(2, 5). Find C. Then find C when A(0, 0), B(a, 0) and D(b, c).
ABCD is a parallelogram with A(0, 0), B(13, 0) and D(5, 12). Show that AD = AB, so it is a rhombus, and find C.

Proving with variables

Use the square O(0, 0), P(a, 0), Q(a, a), R(0, a) to prove that the diagonals of a square are congruent and perpendicular.
Use the rectangle (0, 0), (2a, 0), (2a, 2b), (0, 2b) to prove that the midpoints of the sides of a rectangle are the vertices of a rhombus.
An isosceles trapezoid is placed symmetrically about the y-axis: A(-a, 0), B(a, 0), C(b, c), D(-b, c), with a > 0, b > 0, c > 0. Prove that its diagonals are congruent.
A trapezoid has A(0, 0), B(2a, 0), C(2b, 2c), D(2d, 2c) with a > 0, c > 0 and d < b. M and N are the midpoints of the legs \overline{AD} and \overline{BC}. Prove that \overline{MN} is parallel to the bases and that \displaystyle MN = \frac{1}{2}(AB + DC).
A quadrilateral has vertices A(0, 0), B(2a, 0), C(2b, 2c), D(2d, 2e). Prove that the midpoints P, Q, R, S of \overline{AB}, \overline{BC}, \overline{CD}, \overline{DA} are the vertices of a parallelogram.
A triangle has A(0, 0), B(6a, 0), C(6b, 6c). Show that the point G(2a + 2b, 2c) is two thirds of the way from each vertex to the midpoint of the opposite side, so all three medians pass through it.
Prove that a point (x, y) is the same distance from (-a, 0) and (a, 0), where a > 0, exactly when x = 0.

Deciding with coordinates

A parallelogram has A(0, 0), B(a, 0), D(b, c), C(a + b, c) with a > 0, c > 0. Show that its diagonals are perpendicular exactly when b^2 + c^2 = a^2, that is, exactly when it is a rhombus.
A triangle has A(0, 0), B(2a, 0), C(2b, 2c) with a > 0, c > 0. For which b is the median from C perpendicular to \overline{AB}? Show that this happens exactly when CA = CB.
Which rectangles (0, 0), (a, 0), (a, b), (0, b) have perpendicular diagonals?
A triangle has A(0, 0), B(2a, 0), C(2b, 2c) with a > 0, c > 0, and M is the midpoint of \overline{AB}. Show that MC = MA exactly when the angle at C is a right angle, the converse of Example 2.

Reasoning

Explain why a proof about all triangles may begin “place A at (0, 0), B at (a, 0) with a > 0, and C at (b, c) with c > 0” without losing any triangle.
A student proves that the diagonals of a parallelogram bisect each other using A(0, 0), B(a, 0), C(a, b), D(0, b). Every computation is correct. What has been proved, and what should the placement be?
Another student places a parallelogram at (0, 0), (a, 0), (a + b, b), (b, b) with a > 0, b > 0. Count the letters, and say which parallelograms this placement covers.
Explain why the vertices are written (2a, 0) and (2b, 2c) when midpoints are needed, and why nothing is lost by doing so.
Explain why coordinate proofs are well suited to theorems about parallel and perpendicular sides but poorly suited to a theorem such as “the angles of a triangle add to 180^\circ”.

Answers

P(0, 0), Q(s, 0), R(s, s), S(0, s), and \displaystyle PR = s\sqrt{2}, since \displaystyle PR = \sqrt{s^2 + s^2} = \sqrt{2s^2}.
Y(-a, 0), Z(a, 0), X(0, h), and \displaystyle XY = XZ = \sqrt{a^2 + h^2}. X is on the y-axis because it is equally far from Y and Z, as in Example 1(b).
\displaystyle C(a, a\sqrt{3}). CA = CB puts C on the line x = a, as in Example 1(b). Then CA = 2a gives a^2 + y^2 = 4a^2, so y^2 = 3a^2 and, with y > 0, \displaystyle y = a\sqrt{3}.
C(9, 5), and in general C(a + b, c). C is level with D because \overline{DC} is parallel to the horizontal \overline{AB}, and \overline{BC} \parallel \overline{AD} makes the run from B to C equal the run from A to D, as in Example 3.
\displaystyle AD = \sqrt{25 + 144} = 13 = AB, and C(18, 12), from C(a + b, c) with a = 13, b = 5, c = 12.
\displaystyle OQ = \sqrt{a^2 + a^2} = a\sqrt{2} and \displaystyle PR = \sqrt{(0 - a)^2 + a^2} = a\sqrt{2}, so they are congruent. The slope of \overline{OQ} is \displaystyle \frac{a}{a} = 1 and of \overline{PR} is \displaystyle \frac{a - 0}{0 - a} = -1; the product is -1, so they are perpendicular.
The midpoints are (a, 0), (2a, b), (a, 2b), (0, b). Each side of the inner quadrilateral has run a and rise b in size, so each has length \displaystyle \sqrt{a^2 + b^2}. A quadrilateral with four equal sides is a rhombus.
\displaystyle AC = \sqrt{(b + a)^2 + c^2} and \displaystyle BD = \sqrt{(-b - a)^2 + c^2} = \sqrt{(a + b)^2 + c^2}, so AC = BD.
M = (d, c) and N = (a + b, c). Both have y-coordinate c, so \overline{MN} is horizontal, parallel to the horizontal bases (c > 0 keeps it off the x-axis). MN = a + b - d, which is positive since a > 0 and d < b, and AB + DC = 2a + (2b - 2d), half of which is a + b - d.
P(a, 0), Q(a + b, c), R(b + d, c + e), S(d, e). From P to Q the run is b and the rise c; from S to R the run is b and the rise c. From Q to R the run is d - a and the rise e; from P to S the same. Equal runs and rises give equal slopes, or two vertical sides when the run is 0, so both pairs of opposite sides are parallel: PQRS is a parallelogram.
The point two thirds of the way from X to Y is \displaystyle X + \frac{2}{3}(Y - X). From A to the midpoint (3a + 3b, 3c) of \overline{BC}: (2a + 2b, 2c). From B(6a, 0) to the midpoint (3b, 3c) of \overline{AC}: (6a + 2b - 4a,\ 2c) = (2a + 2b, 2c). From C(6b, 6c) to the midpoint (3a, 0) of \overline{AB}: (6b + 2a - 4b,\ 6c - 4c) = (2a + 2b, 2c). All three give G.
Squaring the equal distances, (x + a)^2 + y^2 = (x - a)^2 + y^2, which simplifies to 4ax = 0. Since a > 0, that holds exactly when x = 0. Every step can be reversed, so both directions are proved: the y-axis is the perpendicular bisector of the segment.
The product of the slopes is \displaystyle \frac{c}{a + b} \cdot \frac{c}{b - a} = \frac{c^2}{b^2 - a^2}, and this is -1 exactly when c^2 = a^2 - b^2, that is b^2 + c^2 = a^2, which says AD = AB. If b = a or b = -a one diagonal is vertical and the other is not horizontal (as c > 0), so they are not perpendicular; and b^2 + c^2 = a^2 fails too, since it would need c = 0.
\displaystyle b = \frac{a}{2}. The midpoint of \overline{AB} is (a, 0), and the median from C(2b, 2c) to it is perpendicular to the horizontal \overline{AB} exactly when it is vertical, 2b = a. And CA = CB means 4b^2 + 4c^2 = (2a - 2b)^2 + 4c^2, which simplifies to 8ab = 4a^2, so 2b = a since a > 0: the same condition.
Exactly the squares, a = b. The slopes are \displaystyle \frac{b}{a} and \displaystyle -\frac{b}{a}, with product \displaystyle -\frac{b^2}{a^2}, which is -1 exactly when b^2 = a^2, and a, b > 0.
M = (a, 0). MC = MA means (2b - a)^2 + 4c^2 = a^2, which simplifies to b^2 - ab + c^2 = 0. The angle at C is right when the slopes \displaystyle \frac{c}{b} of \overline{CA} and \displaystyle \frac{c}{b - a} of \overline{CB} multiply to -1: c^2 = -b(b - a), which is again b^2 - ab + c^2 = 0. If b = 0 or b = a one side is vertical and the other not horizontal, so the angle is not right, and b^2 - ab + c^2 = c^2 > 0 agrees.
Sliding, turning and flipping are rigid motions: they change no length or angle, so they keep midpoints, parallel lines and perpendicular lines. Slide A to the origin, turn until B is on the positive x-axis, so B = (a, 0) with a = AB > 0, and flip across the x-axis if C is below it. C cannot lie on the axis, or the triangle would be flat, so c > 0. b is left free, so triangles with any angle at A are included.
Only that the diagonals of a rectangle bisect each other, since the placement has right angles at every vertex and two letters where a parallelogram needs three. The general placement is A(0, 0), B(a, 0), C(a + b, c), D(b, c), as in Example 3.
Two letters, where a parallelogram needs three. The side from (0, 0) to (b, b) has slope 1, so it makes a 45^\circ angle with the x-axis: the placement covers only parallelograms with a 45^\circ angle (and a 135^\circ angle beside it).
The midpoint formula divides each sum by 2. With doubled coordinates every sum is even, so the midpoints have no fractions, and the later distances and slopes stay short. Nothing is lost because every number is twice another number: as a ranges over all positive numbers, so does 2a.
Parallel and perpendicular are read straight off slopes (equal, or product -1), and lengths and midpoints come from short formulas, so those theorems become algebra on the letters. The measure of a general angle has no such formula in coordinates; it needs trigonometry, and the sum of three such angles does not simplify to 180^\circ by algebra alone. A synthetic proof with parallel lines is far shorter.