Geometry · G.CO.D.12

Why a construction works

A compass and a straightedge have produced a line that looks perpendicular, or a ray that looks as if it cuts an angle in half. How do you know it really does, every time, and not just in this one drawing?

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Why a construction works

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Why this exists

A drawing shows one case. Bisect a segment with compass and straightedge, measure the two halves with a ruler, and they agree to the width of a pencil line. That is evidence about one segment of one length, and it is no more exact than the ruler that read it. A construction promises more: that its steps give the midpoint of every segment, exactly.

A promise about every case needs a proof. The steps of a construction create facts about distances and lines, and the result is a claim about the finished figure, such as “this line is perpendicular to \overline{AB} and passes through its midpoint”. Justifying the construction means connecting the two with the tools of proof: each fact the drawing creates becomes a given, and congruent triangles, the perpendicular bisector or a reflection carry the givens to the conclusion.

The payoff is trust. Once a construction has been justified, later constructions and later proofs can use “draw the perpendicular bisector” as a single step, because it has been shown to work in every case.

Definition

A justification of a construction is a proof that the figure its steps produce has the property the construction claims, for every allowed starting figure. Its givens are the facts the two tools create, and nothing else.

An arc gives equal distances. Every point of an arc drawn about a center C with compass opening r is at distance r from C, because a circle is exactly the set of points at distance r from its center. Two arcs drawn without changing the opening have the same radius.

A crossing lies on both arcs. Where two arcs meet, the point has both distances at once.

The straightedge draws the one line. Through two different points there is exactly one line, and the straightedge draws it.

A fold or a geometry program standing in for the two tools brings its own guaranteed facts, set out under Results.

Lengths and angle measures read from the drawing are never givens. A justification that says “the angle measures 90^\circ on my protractor” has justified nothing, however careful the drawing.

From these givens the ordinary tools of geometry finish the job: SSS and SAS, corresponding parts of congruent triangles, the perpendicular bisector of a segment, the properties of a rhombus, and the definitions of the rigid motions.

Intuition

The habit that opens almost every justification is to mark the equal radii. Once the construction is drawn, join each center to each point found with it, and put matching tick marks on segments drawn with the same compass opening. The congruent triangles are then usually in plain sight.

In bisecting a segment, arcs of one radius about A and about B cross at P and Q. Joining the centers to the crossings gives four segments of that radius, and two triangles, \triangle PAQ and \triangle PBQ, that share the side \overline{PQ}.

Figure 1Mark the equal radii. The four ticked segments were all drawn with one compass opening, and the triangles they make do the proving.

With the ticks drawn the figure is a rhombus PAQB, or a kite when the arcs above and below the segment use different openings, and the facts needed are facts about that shape: its diagonal \overline{PQ} cuts the other diagonal \overline{AB} in half, at right angles. Most of the classic constructions hide a rhombus or a kite in the same way, which is why their justifications sound so alike.

Results

Two facts that do most of the work

Equidistant points. If PA = PB, then P lies on the perpendicular bisector of \overline{AB}. This is the converse of the perpendicular bisector theorem. Two different points with this property fix the line:

\begin{aligned} &PA = PB \text{ and } QA = QB,\ P \ne Q \\ &\implies \overleftrightarrow{PQ} \text{ is the perpendicular bisector of } \overline{AB} \end{aligned}

The kite. If XA = XB and YA = YB, with X \ne Y, then \triangle XAY \cong \triangle XBY by SSS, so \angle AXY \cong \angle BXY and \angle AYX \cong \angle BYX.

Constructions with folding and software

A fold is a reflection. Folding flat paper so that a point A lands on a point B makes a crease that is the perpendicular bisector of \overline{AB}. Folding an angle so that one arm lies along the other makes a crease that bisects the angle.

A geometry program is a compass and straightedge. In dynamic geometry software, the circle tool (a center and a point on the circle, or a center and a copied radius) is the compass, and the line tool through two points is the straightedge. A point made as the crossing of two objects stays on both of them; a point dropped by eye is tied to nothing. That gives the drag test: move the given points, and a true construction keeps its property in every position, because its justification never used any particular size. A figure that only looked right comes apart as soon as it is dragged. The justification of a software construction is the same proof as on paper.

Why it is true

Equidistant points. Let M be the midpoint of \overline{AB}, and suppose PA = PB.

If P is on the line AB, it is a point of that line equally far from A and B. A point outside the segment is nearer one end than the other, and for a point inside it the two distances add to AB, so each is \displaystyle \frac{AB}{2}. So P = M, which is on the perpendicular bisector.

Otherwise \triangle PAM and \triangle PBM have PA = PB, AM = BM and the common side \overline{PM}, so they are congruent by SSS, and \angle PMA \cong \angle PMB. The two angles form a linear pair, so they add to 180^\circ, and being equal each is 90^\circ. So \overleftrightarrow{PM} meets \overline{AB} at its midpoint at right angles: it is the perpendicular bisector, and P is on it.

A second point Q with QA = QB is on the same line. A segment has exactly one perpendicular bisector, and two different points of a line fix it, so \overleftrightarrow{PQ} is that line.

The kite. XA = XB, YA = YB, and \overline{XY} is a side of both triangles, so \triangle XAY \cong \triangle XBY by SSS, with A matched to B. Corresponding parts of Congruent triangles are congruent, so \angle AXY \cong \angle BXY and \angle AYX \cong \angle BYX.

The fold. A fold presses each point on one side of the crease onto a point on the other side, keeps every distance, and leaves the points of the crease where they are: it is the reflection in the crease line. By the definition of a reflection in a line \ell, for a point A off \ell, the line \ell is the perpendicular bisector of the segment from A to its image. The fold sends A to B, so the crease is the perpendicular bisector of \overline{AB}.

For an angle at V folded arm onto arm, the fold carries the ray \overrightarrow{VX} onto \overrightarrow{VY}, so it carries the endpoint V to the endpoint V. A reflection fixes only the points of its line, so V is on the crease. Reflections keep angle measure, and the reflection carries the angle between the crease and \overrightarrow{VX} onto the angle between the crease and \overrightarrow{VY}, so those two angles are equal.

Worked examples

Example 1

Bisecting a segment

AB = 12 cm. Arcs of radius 10 cm are drawn about A and about B; they cross above \overline{AB} at P and below it at Q, and \overleftrightarrow{PQ} meets \overline{AB} at M. Justify that \overleftrightarrow{PQ} is the perpendicular bisector of \overline{AB}, and find MP and PQ.

Solution

The givens come from the arcs. P is on the arc about A and on the arc about B, both of radius 10, so PA = PB = 10. In the same way QA = QB = 10.

First the two triangles that share \overline{PQ}. \triangle PAQ and \triangle PBQ have PA = PB, QA = QB and the common side \overline{PQ}, so \triangle PAQ \cong \triangle PBQ by SSS, and as corresponding parts \angle APQ \cong \angle BPQ.

Next the two triangles that meet at M. P and Q are on opposite sides of \overline{AB}, so M lies on \overline{PQ}, and \angle APM, \angle BPM are the angles just shown congruent. \triangle APM and \triangle BPM have PA = PB, \angle APM \cong \angle BPM and the common side \overline{PM}, so they are congruent by SAS. Hence AM = BM, so M is the midpoint, and \angle AMP \cong \angle BMP. Those two angles form a linear pair, so each is 90^\circ. The line meets \overline{AB} at its midpoint at right angles: it is the perpendicular bisector.

The equidistant points fact gives the same conclusion in one line, since P and Q are each equally far from A and B; the two-triangle argument is that fact proved over again in this one case.

For the lengths, \displaystyle AM = \frac{12}{2} = 6 and \triangle AMP has its right angle at M, so the The Pythagorean theorem gives

MP = \sqrt{10^2 - 6^2} = 8

The same reasoning below the segment gives MQ = 8, and M lies between P and Q, so PQ = 16 cm.

Example 2

Bisecting an angle

\angle XVY measures 72^\circ. An arc about V crosses \overrightarrow{VX} at D and \overrightarrow{VY} at E. Then, with one compass opening s, arcs are drawn about D and about E, crossing at a point F inside the angle. Justify that \overrightarrow{VF} bisects the angle, and give the measures of the two halves.

Solution

The givens: D and E are on one arc about V, so VD = VE; and F is on the arc about D and on the arc about E, both of opening s, so DF = EF = s.

Figure 2Bisecting an angle. \triangle VDF and \triangle VEF have three pairs of equal sides.

Join D and E to F, the dashed segments in the figure. \triangle VDF and \triangle VEF have VD = VE, DF = EF and the common side \overline{VF}, so \triangle VDF \cong \triangle VEF by SSS, with D matched to E. Corresponding angles are congruent, so \angle DVF \cong \angle EVF. These are \angle XVF and \angle YVF, named by other points of the same rays. F is inside the angle, so by the angle addition postulate the two parts make up \angle XVY, and being equal each is half of it:

m\angle XVF = m\angle YVF = \frac{72^\circ}{2} = 36^\circ

When s is the same opening as the first arc, VD = DF = FE = EV and VDFE is a rhombus, and the justification can be put as “a diagonal of a rhombus bisects the angles at its ends”. The SSS argument above is the proof of that property, and it does not need s to equal the first opening.

Example 3

A perpendicular through a point on a line

P is a point on a line \ell. An arc about P crosses \ell at A and B, with PA = 5 cm. With the compass opened to 13 cm, arcs about A and about B cross at Q. Justify that \overleftrightarrow{PQ} \perp \ell, and find PQ.

Solution

One arc about P gives PA = PB = 5, and the two arcs of 13 cm give QA = QB = 13. An arc about a point of a line crosses the line once on each side of the point, so \overrightarrow{PA} and \overrightarrow{PB} are opposite rays.

\triangle QPA and \triangle QPB have QA = QB, PA = PB and the common side \overline{QP}, so they are congruent by SSS, and \angle QPA \cong \angle QPB. Because \overrightarrow{PA} and \overrightarrow{PB} are opposite rays, these two angles form a linear pair and add to 180^\circ; being equal, each is 90^\circ. So \overleftrightarrow{PQ} \perp \ell.

\triangle QPA has its right angle at P, with hypotenuse QA = 13 and leg PA = 5, so

PQ = \sqrt{13^2 - 5^2} = 12

and PQ = 12 cm. Seen another way, this is Example 1 in disguise: the first arc made a segment \overline{AB} whose midpoint is P, and Q is equally far from its ends, so \overleftrightarrow{PQ} is the perpendicular bisector of \overline{AB}.

Example 4

A perpendicular from a point off the line

P is not on the line \ell. With the compass opened to 5 cm, an arc about P crosses \ell at A and B, and AB = 8 cm. Without changing the opening, arcs about A and about B cross at a point Q on the other side of \ell. Justify that \overleftrightarrow{PQ} \perp \ell, and find the distance from P to \ell and the length PQ.

Solution

All four distances PA, PB, QA and QB equal the one opening, 5 cm. So P and Q are two different points, each equally far from A and B, and by the equidistant points fact \overleftrightarrow{PQ} is the perpendicular bisector of \overline{AB}. That segment lies on \ell, so \overleftrightarrow{PQ} \perp \ell, meeting it at the midpoint M of \overline{AB}.

Figure 3The perpendicular from a point off the line. P and Q are each equally far from A and B.

Two other justifications reach the same place. With four equal sides, PAQB is a rhombus, and the diagonals of a rhombus are perpendicular bisectors of each other, so \ell also cuts \overline{PQ} in half. With a rigid motion: the reflection in \ell fixes A and B and keeps distances, so it sends P to a point 5 cm from both A and B on the other side of \ell. The two circles of radius 5 about A and B meet only at P and Q, so that point is Q: Q is the mirror image of P, and \ell is the perpendicular bisector of \overline{PQ}.

For the lengths, AM = 4 and the angle at M is right, so

PM = \sqrt{5^2 - 4^2} = 3

\overline{PM} is perpendicular to \ell, so the distance from P to \ell is 3 cm. Since QM = PM = 3, PQ = 6 cm.

Example 5

A parallel through a point

P is not on the line \ell. A line t is drawn through P, crossing \ell at A; R is a point of \ell and T a point of t beyond P, so that P is between A and T, and m\angle PAR = 58^\circ. The angle \angle PAR is copied at P, with one arm along \overrightarrow{PT} and the new arm \overrightarrow{PS} on the same side of t as R. Justify that \overleftrightarrow{PS} \parallel \ell, and find m\angle APS.

Solution

The angle copy was justified by SSS: it gives \angle TPS \cong \angle PAR, so m\angle TPS = 58^\circ.

Read the two angles as angles made by the transversal t. \angle PAR lies at A, between t and \ell, on the side of t containing R and on the side of \ell containing T. \angle TPS lies at P, between t and \overleftrightarrow{PS}, on the same side of t and on the side of \overleftrightarrow{PS} containing T. They sit in matching positions, so they are corresponding angles, and they are congruent. By the converse of the corresponding angles postulate (Proving lines parallel), \overleftrightarrow{PS} \parallel \ell.

\overrightarrow{PA} and \overrightarrow{PT} are opposite rays, so \angle APS and \angle TPS form a linear pair:

m\angle APS = 180^\circ - 58^\circ = 122^\circ

As a check, \angle APS and \angle PAR are same side interior angles, and 122^\circ + 58^\circ = 180^\circ, as parallel lines require. Since only one line through P is parallel to \ell, the construction has found it.

Example 6

Finding a point by folding

A rectangular sheet ABCD is 8 cm wide and 4 cm tall, with A = (0, 0), B = (8, 0), C = (8, 4) and D = (0, 4). It is folded so that A lands on C, and the crease meets \overline{AB} at E and \overline{DC} at G. Justify that EA = EC, then find E, G and the length of the crease.

Solution

The fold is a reflection that sends A to C, so the crease is the perpendicular bisector of \overline{AC}. E is on the crease, and every point of the perpendicular bisector is equally far from the ends of the segment, so EA = EC. The fold has found, without any measuring, the point of \overline{AB} as far from C as from A.

Figure 4Folding A onto C. The crease is the perpendicular bisector of \overline{AC}, so E is as far from C as from A.

To locate it, use coordinates. The midpoint of \overline{AC} is (4, 2). The slope of \overline{AC} is \displaystyle \frac{4}{8} = \frac{1}{2}, and perpendicular slopes multiply to -1, so the crease has slope -2:

y - 2 = -2(x - 4)

Setting y = 0 gives x - 4 = 1, so E = (5, 0) and AE = 5 cm. Setting y = 4 gives x - 4 = -1, so G = (3, 4). The crease runs from E to G, with length

EG = \sqrt{(5 - 3)^2 + 4^2} = \sqrt{20} = 2\sqrt{5} \text{ cm}

As a check that uses only the justified fact, call AE = x. Then EB = 8 - x, and the right triangle \triangle EBC gives EC^2 = (8 - x)^2 + 4^2. Setting EC = EA, (8 - x)^2 + 16 = x^2, so 80 - 16x = 0 and x = 5, as before. In a geometry program the same point comes from the perpendicular bisector tool and an intersection, and dragging C to reshape the rectangle keeps EA = EC in every position.

Example 7

Finding the gap in a construction

Lena bisects a right angle \angle XVY. Her first arc, of radius 4, gives D on \overrightarrow{VX} and E on \overrightarrow{VY}. She sets the compass to 5 for the arc about D, but it slips, and the arc about E is drawn at 7. The two arcs cross at F inside the angle. Does \overrightarrow{VF} bisect the angle? Where does the justification break?

Solution

Put V at the origin with \overrightarrow{VX} along the positive x-axis and \overrightarrow{VY} along the positive y-axis, so D = (4, 0) and E = (0, 4). The point F satisfies both circle equations:

(x - 4)^2 + y^2 = 25 \qquad x^2 + (y - 4)^2 = 49

Subtracting the first from the second, 8x - 8y = 24, so y = x - 3. Substituting into the first, (x - 4)^2 + (x - 3)^2 = 25, which is 2x^2 - 14x = 0, so x = 0 or x = 7. The point (0, -3) is outside the angle, so F = (7, 4). As a check, \displaystyle DF = \sqrt{3^2 + 4^2} = 5 and EF = 7.

Figure 5Unequal arcs from D and E. The ray through F is not the bisector, and the proof shows exactly why.

\overrightarrow{VF} makes with \overrightarrow{VX} the angle whose tangent is \displaystyle \frac{4}{7}, about 29.7^\circ, and with \overrightarrow{VY} about 60.3^\circ. It does not bisect the angle; the bisector of a right angle placed this way is the line y = x.

The proof of Example 2 breaks at the SSS step. \triangle VDF and \triangle VEF have VD = VE and the common side \overline{VF}, but DF = 5 and EF = 7: only two pairs of sides match, and the triangles are not congruent. The equality DF = EF comes only from drawing both arcs with one opening. With it, V and F are both equally far from D and E, and the equidistant points fact puts F on the perpendicular bisector of \overline{DE}, the line y = x.

Boundaries

The arcs have to cross. Two circles of radius r whose centers are d apart meet at two points only when d < 2r: a common point is r from each center, so by the Triangle inequality d \le 2r, with equality only when the point is the midpoint of the centers. So bisecting \overline{AB} needs an opening greater than \displaystyle \frac{AB}{2}; at exactly \displaystyle \frac{AB}{2} the arcs touch only at the midpoint and give no line. In the same way, the arcs about D and E in an angle bisection need an opening greater than \displaystyle \frac{DE}{2}, and the arcs about A and B in Example 3 an opening greater than PA.

Choose the right crossing. When the arcs about D and E in an angle bisection are drawn with the same opening as the first arc, one of their two crossings is V itself, since V is that far from both. F must be the other one. Both crossings of equal arcs about D and E lie on the bisecting line, but a crossing on the far side of V lies on the opposite ray, and the ray drawn from V must point into the angle.

Equal openings where the proof uses them. Each justification rests on particular equalities, and the steps that create them must keep the compass fixed, as Example 7 shows. Openings the proof never uses are free: the first arc of an angle bisection can have any radius.

A justification is about the ideal figure. It proves that the exact construction is exact. A pencil line has width and a compass can slip; those are errors of drawing, not of the construction, and they are also why measuring a drawing can never replace the proof.

Folding needs a flat sheet and an exact fold. The crease is a straight line, and the fold a reflection, only for flat paper folded so that the point really lands on its target.

Software constrains only what was constructed. A point dropped near a crossing, rather than on it, moves freely when the figure is dragged, and any property it seemed to have was never given.

A justification checks a construction; it cannot supply one. Some figures have no compass and straightedge construction at all. Trisecting a general angle with these two tools alone was proved impossible in the nineteenth century, so no justification of such a trisection can exist.

Where this goes next

The two facts here carry the remaining constructions. Two perpendicular bisectors of a triangle’s sides meet at a point equally far from all three vertices, which is how the circumscribed circle is found and justified, and the angle bisectors of a triangle meet at the center of the inscribed circle. The angle copy and the corresponding angles converse justify perpendicular and parallel lines through a point; arcs of two given lengths are justified the same way in constructing a triangle from given parts; and the equilateral triangle, square and hexagon and dividing a segment into equal parts each rest on a short proof of the same kind.

The equidistant points fact is also the first example of a locus, a set of points described by a distance condition, and the fold as a reflection continues into Line symmetry.

Exercises

Lengths a construction guarantees

\overline{AB}, with AB = 10 cm, is bisected using arcs of radius 13 cm about A and B, which cross at P and Q; \overleftrightarrow{PQ} meets \overline{AB} at M. Find MP and PQ.
The same segment, AB = 10 cm, is bisected with the compass opened to 4 cm. What goes wrong, and which openings work?
An angle of 128^\circ is bisected, and then one of the two halves is bisected again. Give the measures of the angles produced.
An angle of 60^\circ at V is bisected with every arc at the same opening, 6 cm: the first arc gives D and E, and the arcs about D and E cross at F inside the angle. Explain why VDFE is a rhombus, and find DE and VF.
In constructing the perpendicular through a point P on a line, the first arc gives PA = PB = 8 cm, and the arcs about A and B are drawn at 17 cm, crossing at Q. Find PQ.
A point P is 9 cm from a line \ell. The perpendicular from P is constructed as in Example 4 with the compass at 15 cm throughout, giving A and B on \ell and Q on the other side. Find AB and PQ.
On a coordinate grid, A = (1, 2) and B = (7, 10). Arcs of radius 13 about A and B cross at P, to the upper left of \overline{AB}, and at Q, to the lower right. Find the coordinates of P and Q.

Which fact does the work

In Example 3, name the two triangles shown congruent, the criterion used, and the fact that turns two congruent angles into right angles.
In the parallel construction of Example 5, suppose instead m\angle PAR = 47^\circ. Find m\angle APS, and name the theorem that makes \overleftrightarrow{PS} \parallel \ell.
When every arc of an angle bisection uses one opening, the arcs about D and E cross at two points. Name them, say which one is used, and explain why the figure is a rhombus whose diagonal bisects the angle.
In Example 4, which rigid motion carries P to Q? Use it to explain why \ell is the perpendicular bisector of \overline{PQ}.

Folding and software

A square sheet of side 10 cm is folded so that one corner lands on the opposite corner. Where is the crease, why, and how long is it?
A rectangular sheet is 8 cm wide and 6 cm tall, with corners A = (0, 0) and C = (8, 6) opposite each other. It is folded so that A lands on C. Find where the crease meets the bottom edge and the top edge, and how long the crease is.
An angle of 104^\circ is drawn on paper and folded so that one arm lies along the other. What angle does the crease make with each arm? The sheet is opened, and the angle between the crease and one arm is folded in half the same way. What angle does the new crease make with that arm?
In a geometry program, Tomas draws \overline{AB}, drops a point P that looks equally far from A and B, and draws a line through P that looks perpendicular to \overline{AB}. When he drags B, the line no longer bisects the segment. Explain what the drag shows and what a construction does differently.
In a geometry program, Rosa draws the circle with center A through B and the circle with center B through A, where AB = 6, and joins their two crossing points. Explain why her line is the perpendicular bisector of \overline{AB}, and find the distance between the two crossing points.

Finding the gap

An arc of radius 5 about A = (0, 0) and an arc of radius 7 about B = (8, 0) cross at two points. Find them. Is the line through them perpendicular to \overline{AB}? Does it bisect \overline{AB}?
Jun is bisecting a right angle at V = (0, 0), but his first arc slipped: it gave D = (4, 0) on one arm and E = (0, 6) on the other. His next two arcs had equal radius and crossed at F = (8, 7). Check that DF = EF, and decide whether \overrightarrow{VF} bisects the angle.
In constructing the perpendicular through a point P on a line, Ama draws the arcs about A and B with the same opening as the first arc, PA. What happens, and why?

Reasoning

Justify the angle bisection a second way: show that V and F are both equally far from D and E, deduce that \overleftrightarrow{VF} is the perpendicular bisector of \overline{DE}, and then explain why \overrightarrow{VF} bisects \angle DVE.
Prove that the construction of Example 4 still gives a perpendicular when the arcs about A and B use a different opening from the first arc, as long as they cross on the other side of \ell. What is lost?
Prove that when a sheet is folded so that A lands on B, every point of the crease is equally far from A and from B.
Explain why a justification must never use a length or an angle measured from the drawing, however accurate the drawing is.

Answers

MP = 12 cm and PQ = 24 cm. AM = 5 and the angle at M is right, so \displaystyle MP = \sqrt{13^2 - 5^2} = 12; below the segment MQ = 12 likewise, and PQ = 12 + 12 = 24.
The arcs do not meet. A point on both would be 4 from A and 4 from B, and then by the triangle inequality AB \le 4 + 4 = 8, but AB = 10. Any opening greater than 5 cm, half of AB, works. At exactly 5 cm the arcs touch only at the midpoint, which gives no line.
64^\circ and 64^\circ, then 32^\circ and 32^\circ: \displaystyle \frac{128^\circ}{2} = 64^\circ, and \displaystyle \frac{64^\circ}{2} = 32^\circ.
DE = 6 cm and \displaystyle VF = 6\sqrt{3} cm. All four sides VD, VE, DF and EF were drawn at 6 cm, so VDFE is a rhombus. \triangle VDE has VD = VE, so its base angles are equal, and they add to 180^\circ - 60^\circ = 120^\circ; all three angles are 60^\circ, the triangle is equilateral, and DE = 6. The diagonals of a rhombus are perpendicular bisectors of each other, so with M the midpoint of \overline{DE}, \displaystyle VM = \sqrt{6^2 - 3^2} = 3\sqrt{3} and \displaystyle VF = 2 \cdot 3\sqrt{3} = 6\sqrt{3}.
PQ = 15 cm. The construction makes \angle QPA a right angle, so \displaystyle PQ = \sqrt{17^2 - 8^2} = 15.
AB = 24 cm and PQ = 18 cm. \overleftrightarrow{PQ} meets \ell at the midpoint M of \overline{AB} at right angles, with PM = 9, so \displaystyle AM = \sqrt{15^2 - 9^2} = 12 and AB = 24. Q is the mirror image of P in \ell, so PQ = 2 \cdot 9 = 18.
\displaystyle P = \left(-\frac{28}{5}, \frac{66}{5}\right) and \displaystyle Q = \left(\frac{68}{5}, -\frac{6}{5}\right). \displaystyle AB = \sqrt{6^2 + 8^2} = 10 and the midpoint is M = (4, 6), so \displaystyle MP = MQ = \sqrt{13^2 - 5^2} = 12. \overline{AB} points along \displaystyle \left(\frac{3}{5}, \frac{4}{5}\right), so the perpendicular points along \displaystyle \left(-\frac{4}{5}, \frac{3}{5}\right), and \displaystyle P = (4, 6) + 12\left(-\frac{4}{5}, \frac{3}{5}\right). Check: \displaystyle PA^2 = \left(\frac{33}{5}\right)^2 + \left(\frac{56}{5}\right)^2 = 169.
\triangle QPA \cong \triangle QPB by SSS, from QA = QB, PA = PB and the shared side \overline{QP}. The congruent angles \angle QPA and \angle QPB form a linear pair, so they add to 180^\circ, and equal angles adding to 180^\circ are each 90^\circ.
m\angle APS = 133^\circ; the converse of the corresponding angles postulate. The copy gives m\angle TPS = 47^\circ, corresponding to \angle PAR, so the lines are parallel, and \angle APS forms a linear pair with \angle TPS: 180^\circ - 47^\circ = 133^\circ.
V and F. VD = VE equals the one opening, so V is on both arcs; the other crossing, F, is the one used, since a line needs a second point. Then VD = VE = DF = EF, four equal sides, so VDFE is a rhombus. By SSS, \triangle VDF \cong \triangle VEF, so \angle DVF \cong \angle EVF: the diagonal \overline{VF} bisects the angle at V.
The reflection in \ell. It fixes A and B, which are on \ell, and keeps distances, so the image of P is 5 cm from A and from B and lies on the other side of \ell. The circles of radius 5 about A and B meet only at P and Q, so the image is Q. By the definition of a reflection, \ell is the perpendicular bisector of the segment joining a point to its image, which is \overline{PQ}.
Along the other diagonal, \displaystyle 10\sqrt{2} cm long. The fold sends one corner A to the opposite corner C, so the crease is the perpendicular bisector of \overline{AC}. The other two corners are each 10 cm from A and from C, so both lie on that perpendicular bisector, and the crease is the diagonal through them, of length \displaystyle \sqrt{10^2 + 10^2} = 10\sqrt{2}.
Bottom edge at \displaystyle \left(\frac{25}{4}, 0\right), top edge at \displaystyle \left(\frac{7}{4}, 6\right), crease \displaystyle \frac{15}{2} cm long. The crease is the perpendicular bisector of \overline{AC}: it passes through the midpoint (4, 3) with slope \displaystyle -\frac{4}{3}, since \overline{AC} has slope \displaystyle \frac{3}{4}. So \displaystyle y - 3 = -\frac{4}{3}(x - 4); y = 0 gives \displaystyle x = \frac{25}{4} and y = 6 gives \displaystyle x = \frac{7}{4}. The length is \displaystyle \sqrt{\left(\frac{9}{2}\right)^2 + 6^2} = \frac{15}{2}.
52^\circ with each arm, then 26^\circ. The fold is a reflection, which keeps angle measure, so the crease splits the angle into two equal parts, \displaystyle \frac{104^\circ}{2} = 52^\circ; folding the 52^\circ angle the same way gives 26^\circ.
Tomas’s point and line were placed by eye, so nothing ties PA to PB or the line to a right angle; when B moves, P and the line stay where they were and the property disappears. A construction makes P and Q the crossings of circles of equal radius about A and B. The program keeps them on those circles whatever is dragged, so PA = PB and QA = QB stay true, and \overleftrightarrow{PQ} stays the perpendicular bisector, because the justification never used any particular length.
\displaystyle 6\sqrt{3}. Each crossing is on the circle about A through B, so it is AB = 6 from A, and on the circle about B through A, so it is 6 from B. Both crossings are equally far from A and B, so the line through them is the perpendicular bisector. Each crossing is \displaystyle \sqrt{6^2 - 3^2} = 3\sqrt{3} from the midpoint, on opposite sides, so they are \displaystyle 6\sqrt{3} apart.
\displaystyle \left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right) and \displaystyle \left(\frac{5}{2}, -\frac{5\sqrt{3}}{2}\right); perpendicular, yes; bisecting, no. Subtracting x^2 + y^2 = 25 from (x - 8)^2 + y^2 = 49 gives -16x + 64 = 24, so \displaystyle x = \frac{5}{2}, and then \displaystyle y^2 = 25 - \frac{25}{4} = \frac{75}{4}. The line \displaystyle x = \frac{5}{2} is perpendicular to \overline{AB}; in fact A and B are each equally far from the two crossings, so line AB is the perpendicular bisector of the segment joining them. But it meets \overline{AB} at \displaystyle \left(\frac{5}{2}, 0\right), not at the midpoint (4, 0): unequal radii keep the right angle and lose the midpoint.
\displaystyle DF = EF = \sqrt{65}, but \overrightarrow{VF} does not bisect the angle. DF^2 = 4^2 + 7^2 = 65 and EF^2 = 8^2 + 1^2 = 65. \overrightarrow{VF} makes the angle whose tangent is \displaystyle \frac{7}{8}, about 41.2^\circ, with \overrightarrow{VD}, and about 48.8^\circ with \overrightarrow{VE}. The SSS step needs VD = VE as well, and 4 \ne 6; F is on the perpendicular bisector of \overline{DE}, but V is not.
The arcs touch at P and nowhere else. A and B are each PA from P, on opposite sides, so AB = 2 \cdot PA, and two circles of radius PA whose centers are 2 \cdot PA apart meet only at the midpoint of the centers, which is P. There is no second point to draw the line through; the opening must be greater than PA.
The first arc gives VD = VE and the equal arcs give FD = FE, so V and F are two different points each equally far from D and E. By the equidistant points fact, \overleftrightarrow{VF} is the perpendicular bisector of \overline{DE}, meeting it at its midpoint M. \triangle VMD and \triangle VME have VD = VE, MD = ME and the shared side \overline{VM}, so they are congruent by SSS and \angle DVM \cong \angle EVM. M lies on \overline{DE}, inside the angle, and so does F, so \overrightarrow{VM} and \overrightarrow{VF} are the same ray, and \overrightarrow{VF} bisects \angle DVE.
Let the arcs about A and B have opening s, crossing at Q on the other side of \ell. The single arc about P still gives PA = PB, and the equal arcs give QA = QB = s. P and Q are different points, each equally far from A and B, so \overleftrightarrow{PQ} is the perpendicular bisector of \overline{AB}, which lies on \ell; hence \overleftrightarrow{PQ} \perp \ell. What is lost is the rhombus: PAQB is now a kite, Q is no longer the mirror image of P, and \ell need not cut \overline{PQ} in half.
The fold is the reflection in the crease, and it sends A to B. Let X be any point of the crease. A reflection fixes the points of its line, so X goes to X, and it keeps distances, so the distance from X to A equals the distance from the image of X to the image of A, which is XB. So XA = XB.
A construction claims its result for every starting figure, and a drawing is one case; a fact measured from it says nothing about the others. A measurement is also approximate, while the claim is exact: a protractor cannot tell 90^\circ from 90.01^\circ, so a reading of 90^\circ is consistent with an angle that is not right. Only the facts the tools guarantee, equal radii, crossings and the line through two points, hold exactly in every case, so they are the only givens.