Segment length, addition and congruence
A point cuts a segment in two, and somewhere else two segments sit apart, pointing different ways. How do the lengths of the pieces add up to the whole, and when are two separate segments the same size?
Where you are 4 in Segments, rays and length
Why this exists
Lay two sticks end to end in a straight line and the total length is the sum of the two; that is how you measured length in the first place, units laid end to end. In a geometry proof, though, “it looks like they add” is not a reason. Every step has to point at something already agreed. The segment addition postulate is that agreed thing: it says exactly when two lengths add to a third, and it is the fact you cite whenever you split a segment, join two segments into one, or find a piece from the whole.
The second job is comparing segments that do not touch. Look at the two segments in Figure 2 that carry one tick each. One runs gently uphill to the right. The other climbs steeply to the left, from a different starting point. They are not the same segment: their endpoints are four different points. But a ruler gives the same reading on each, and in almost every problem that is all that matters. A builder cutting two table legs, or a proof about the two equal sides of a triangle, needs to say “these two have the same length” while keeping clear that there are two of them. Geometry’s word for that is congruent.
Both ideas have twins for angles: angle addition and congruent angles. There a ray inside an angle splits a measure; here a point between two others splits a length.
Definition
Segments and their lengths
A segment \overline{AB} is the endpoints A and B and every point of the line between them: a figure. AB, with no bar, is its length, a positive number such as 7 or 4.5 cm. Some books write the length as m\overline{AB}, “the measure of segment AB”; it means the same as AB. Naming the ends the other way round changes nothing: \overline{AB} and \overline{BA} are one segment, and AB = BA.
Betweenness and the segment addition postulate
Point B is between A and C when A, B and C are three different points on one line (they are collinear) and B lies on \overline{AC}. In Figure 1, B is between A and C.
Segment Addition Postulate. If B is between A and C, then
In words: when a point sits on a segment, the two shorter pieces it makes add up to the whole segment. The two pieces \overline{AB} and \overline{BC} share the point B, and nothing else.
It is called a postulate because it is accepted as a starting rule, not proved from anything simpler.
Congruent segments
Two segments are congruent when they have the same length.
In symbols, \overline{AB} \cong \overline{CD} means exactly the same thing as
The bar matters here. \cong goes between segments and = goes between lengths. Writing \overline{AB} = \overline{CD} would claim the two are one segment, and \overline{AB} = 5 compares a figure with a number.
On a figure, congruent segments are marked with matching ticks: short strokes across the middle of each segment, the same number on each. In Figure 2, \overline{AB} and \overline{CD} carry one tick each, so \overline{AB} \cong \overline{CD}. \overline{EF} and \overline{GH} carry two each, so \overline{EF} \cong \overline{GH}, and the different number of ticks says this second pair need not match the first. Ticks are information: a figure is not drawn to scale unless it says so, and segments that merely look equal are not given as congruent.
Intuition
Think of a length as a walk along the line. To get from A to C you can walk straight there, or you can stop at B on the way and then carry on. The distance covered is the same either way, so the first leg plus the second leg equals the whole walk. The condition that B is between A and C is what makes B a stop on the way. If B sat beyond C, or off the line altogether, then going to B first would be a detour, and a detour is longer than the straight walk.
A segment’s length is how far apart its two endpoints are. Three things about how it is drawn make no difference to that.
- Where it is. Sliding a segment across the page does not stretch or shrink it.
- Which way it points. Turning it, so that a flat segment stands upright or leans, keeps the same distance between its ends.
- Which end is named first. \overline{AB} and \overline{BA} are one segment.
So congruent segments are exactly the segments you could make fit: trace one on thin paper, slide it and turn it, and it lies along the other, end on end.
A compass does the same job without paper. Set its point on A and its pencil on B, and it holds the length AB. Put the point on C, and the arc it draws passes through every point at that distance from C; \overline{CD} \cong \overline{AB} exactly when D lies on that arc. That is how you copy a segment without reading a ruler.
Results
Segment addition, read three ways. For B between A and C, the postulate gives whichever length you are missing:
Each piece is shorter than the whole. Both pieces have positive length, so AB < AC and BC < AC.
Any number of pieces. If A, B, C and D lie on a line in that order, applying the postulate twice gives
The sum only works for the point in the middle. For three different points A, B, C, the equation AB + BC = AC holds exactly when B is between A and C, and in every other case AB + BC is greater than AC. So the equation can be used backwards, as a test: if the two shorter distances add to the longest, the three points lie on one line with the middle one between the others.
Congruence of segments behaves like equality. For any segments \overline{AB}, \overline{CD} and \overline{EF}:
- Reflexive. \overline{AB} \cong \overline{AB}.
- Symmetric. If \overline{AB} \cong \overline{CD}, then \overline{CD} \cong \overline{AB}.
- Transitive. If \overline{AB} \cong \overline{CD} and \overline{CD} \cong \overline{EF}, then \overline{AB} \cong \overline{EF}.
Each is inherited from the matching property of = between lengths. They are what let a proof cite a shared side, or a chain of congruences.
Same length means one fits on the other. Two segments are congruent exactly when some slide followed by a turn carries one onto the other.
Adding or removing a shared piece. Suppose A, B, C and D lie on a line in that order, as in Figure 3. Then
This is what lets tick marks be passed along: ticks on the two end pieces license ticks on the two long overlapping segments, and back, with no lengths known at all.
Why it is true
The segment addition postulate
A postulate is accepted rather than proved, but the reason this one is sensible to accept is the ruler.
Lay a ruler along the line, so that every point on the line sits at one reading. The length of a segment is then the difference of the readings at its two ends, the larger minus the smaller, as on a number line. Say A reads a and C reads c, with a < c. Because B is between them, its reading b is between them too: a < b < c. Then
and adding them,
So the postulate is what you get by agreeing that measuring a length means reading a scale, the agreement school geometry calls the ruler postulate.
The betweenness test
Suppose B is not between A and C. If B is on the line but beyond C, then C is between A and B, so AB = AC + CB, and
Beyond A works the same way, with AB counted twice. If B is off the line, then A, B, C are the corners of a triangle, and the triangle inequality says that the two sides AB and BC together are longer than the third side AC. In every case the sum is too big, so the only way to get AB + BC = AC exactly is for B to be between. Some axiom systems, such as Anton Petrunin’s, make this test a theorem: there, the fact that AB + BC = AC puts B on \overline{AC} is proved from side-side-side congruence.
Same length means one fits on the other
Take two segments of the same length, \overline{AB} and \overline{CD}.
First slide \overline{AB} so that A lands on C. A slide moves every point the same distance in the same direction, so the length does not change, and the two segments now share an endpoint (Figure 4).
Next turn about C until the moved segment lies along the ray from C through D. A turn about C keeps every point the same distance from C, so B is still AB away from C, and it now sits on that ray. But a ray holds only one point at any given distance from its endpoint, and since AB = CD, that point is D. So B lands on D, and the moved segment lies exactly on \overline{CD}.
Conversely, a slide or a turn never changes a length, so segments that can be carried one onto the other had equal lengths all along.
The shared piece
In Figure 3 the two long segments overlap: both contain the shaded piece \overline{BC}. B is between A and C, so by the segment addition postulate AC = AB + BC. C is between B and D, so BD = BC + CD. Both right sides contain the same BC, so AC = BD exactly when
and, taking BC away from both sides or adding it back, that holds exactly when AB = CD. The definition of congruence turns each equation of lengths into a congruence. Take away the shared piece from two equal wholes and what is left over must match: in Figure 3, AC = BD = 10 and BC = 6, so both outer pieces are 10 - 6 = 4.
Worked examples
Adding two pieces, and finding a missing one
In Figure 5, B is between A and C, with AB = 7 and BC = 12. Find AC. Then, on another line, point Q is between P and R, with PR = 23.4 cm and PQ = 8.9 cm. Find QR.
B is between A and C, so the segment addition postulate applies to these three lengths:
19 is longer than each piece, as a whole must be.
For the second segment the whole is known and a piece is missing. Because Q is between P and R, the postulate still says the two pieces add to the whole, PQ + QR = PR. Put in the two known lengths and subtract 8.9 from both sides:
So QR = 14.5 cm. Check. 8.9 + 14.5 = 23.4.
Pieces written as expressions
In Figure 6, B is between A and C, with AB = 2x + 3, BC = 4x - 1 and AC = 38. Find x and the length of each piece. Then do the same when point Q is between P and R and the whole is an expression too: PQ = 3x - 4, QR = x + 9 and PR = 5x - 7.
The postulate turns the picture into an equation: the two pieces add to the whole.
Now substitute back. AB = 2(6) + 3 = 15 and BC = 4(6) - 1 = 23. Check. 15 + 23 = 38, and both pieces are positive, as lengths must be.
An unknown whole changes nothing: piece plus piece equals whole is still one equation in x.
Substitute x = 12 back into each expression:
Check. 32 + 21 = 53.
Sorting segments on a grid
Which of the five segments in Figure 7 are congruent to each other?
Each segment runs along the grid lines, so its length is the number of squares it spans (see Horizontal and vertical distance), wherever it sits and whichever way it points: AB = 4, CD = 3, EF = 4, GH = 6 and JK = 3.
So \overline{AB} \cong \overline{EF} and \overline{CD} \cong \overline{JK}. \overline{AB} is flat and \overline{EF} stands upright, but turning a segment does not change its length.
\overline{GH}, of length 6, matches nothing. It is twice as long as \overline{CD}, but that is a different relationship from congruent.
Two marked sides of a triangle
In Figure 8 the ticks mark \overline{AB} \cong \overline{AC}. Given AB = 3x + 1, AC = 5x - 9 and BC = 2x + 4, find x, the three side lengths and the perimeter of the triangle.
The ticks are the given information. Congruent segments have equal lengths, so they say AB = AC:
Then AB = 3(5) + 1 = 16 and AC = 5(5) - 9 = 16, which agree. The base carries no tick, so its length comes from its own expression, BC = 2(5) + 4 = 14.
The perimeter is the sum of the three sides, 16 + 16 + 14 = 46.
A triangle with two congruent sides is isosceles; see Isosceles and equilateral triangles.
A shared middle piece
In Figure 9, A, B, C and D lie on a line in that order and \overline{AB} \cong \overline{CD}. Given AB = 2y + 3, CD = 5y - 9 and BC = 6, find y and AC. What can be said about \overline{BD} without adding?
The ticks say the end pieces have equal lengths:
So AB = 2(4) + 3 = 11 and CD = 5(4) - 9 = 11. B is between A and C, so by the segment addition postulate AC = 11 + 6 = 17.
For \overline{BD} no arithmetic is needed. The end pieces carry matching ticks, so the shared-piece result gives \overline{AC} \cong \overline{BD}, and BD = 17 as well. Adding confirms it: BD = 6 + 11 = 17.
A root to throw away
Point B is between A and C, with AB = x^2, BC = 2x + 5 and AC = 20. Find x and both pieces.
SolutionThe postulate gives one equation, and this time it is a quadratic:
So x = -5 or x = 3. The algebra is finished, but the geometry is not: each value has to give real lengths. With x = -5, BC = 2(-5) + 5 = -5, and no segment has a negative length, so that root is thrown away. With x = 3, AB = 3^2 = 9 and BC = 2(3) + 5 = 11.
Check. 9 + 11 = 20, and both pieces are positive.
When two roots both survive
Two congruent segments have lengths x^2 + 2 and 3x + 12. Find every possible length. Then a third segment of length x + 22 is congruent to both; which answer survives?
SolutionCongruent means equal lengths:
So x = 5 or x = -2. Test each, since a length must come out positive:
- x = 5 gives 5^2 + 2 = 27 and 3(5) + 12 = 27.
- x = -2 gives (-2)^2 + 2 = 6 and 3(-2) + 12 = 6.
Both are genuine. Unlike the previous example, a negative x is fine here, because what must be positive is the lengths it produces, not x itself. So the segments could have length 27 or length 6.
The third segment settles it. At x = 5 its length is 5 + 22 = 27, which matches. At x = -2 it is -2 + 22 = 20, which does not match 6. Only x = 5 makes all three congruent, and the common length is 27.
A slanted segment
In Figure 10, A = (1, 1), B = (4, 5), C = (8, 1) and D = (8, 6). Is \overline{AB} \cong \overline{CD}? Then find every point F on the horizontal line through C with \overline{CF} \cong \overline{AB}.
\overline{CD} is upright, so its length is a count of squares: CD = 6 - 1 = 5.
\overline{AB} is slanted, so it cannot be counted. Instead, go from A to B along the grid: 4 - 1 = 3 across and 5 - 1 = 4 up. Those two legs and \overline{AB} form a right triangle, so the Pythagorean theorem gives
Both lengths are 5, so \overline{AB} \cong \overline{CD}, one leaning and one upright.
For F: it must lie on the line y = 1, at distance 5 from C = (8, 1). Going left gives F = (3, 1) and going right gives F = (13, 1): the two places where a compass set to AB and centered on C crosses that line. The same calculation for any two points is the distance formula.
Which point is in the middle?
Three points have PQ = 9, QR = 15 and PR = 6. Do they lie on one line, and if so, which point is between the other two? What changes if instead PR = 8?
SolutionThe postulate, read as a test, says a point is between the other two exactly when its two distances add to the third. If one point is in the middle, the whole is the longest of the three distances, and it joins the two outer points. The longest distance is QR = 15, so the only candidate for the middle is the point not on it, P. Test it:
The sum works, so the three points lie on one line with P between Q and R.
With PR = 8 instead, test all three points, one at a time, as the middle:
- P in the middle needs QP + PR = QR, but 9 + 8 = 17, not 15.
- Q in the middle needs PQ + QR = PR, but 9 + 15 = 24, not 8.
- R in the middle needs PR + RQ = PQ, but 8 + 15 = 23, not 9.
No point is between the other two, so the three points do not lie on one line. They are the corners of a triangle, and 17 > 15 is the triangle inequality at work.
Boundaries
The point has to be between. In the left panel of Figure 11, A, C and B lie on one line in that order, with AC = 4 and CB = 3. Then AB = 7, and AB + BC = 7 + 3 = 10, which is not 4. The postulate still says something true here, but about a different segment: now C is between A and B, so AC + CB = AB.
The points have to be on one line. In the right panel of Figure 11, B is off line AC, with AB = 5, BC = 5 and AC = 8. The two pieces add to 10, not 8: going through B is a detour. Off the line the sum is always too big, which is the triangle inequality.
Lengths have no direction. Because a length is never negative, the postulate needs to know which point is in the middle. Give each piece a plus or minus sign for the way it points along the line, and the signed pieces add for any three points on the line, in any order. That is the directed segment.
Only segments have lengths. A line or a ray goes on for ever, so it has no length to compare. Any line can be slid and turned onto any other line, and likewise any ray onto any ray, so calling two of them congruent would say nothing. Congruence tells you something about segments because a segment stops.
Neither looks nor a ruler can prove congruence. Congruence is exact, and a ruler reading is not. It comes from given information, from matching ticks, or from a reason such as a midpoint or the shared-piece result.
Counting squares works only along the grid. A slanted segment on a grid cannot be measured by counting, and needs the right-triangle argument of Example 8.
Where this goes next
The segment addition postulate is the grammar of every length calculation along a line. Put a point exactly halfway and you have a midpoint, which cuts a segment into two congruent pieces; pass a line through it and you have a segment bisector. Splitting a segment into pieces in a chosen ratio is partitioning a segment, and every point on the perpendicular bisector of \overline{AB} is joined to A and to B by congruent segments.
In a two-column proof the postulate is usually the line with the reason “Segment Addition Postulate”, sitting next to the properties of equality that move the lengths around afterwards. Read as a test it becomes the equality case of the triangle inequality, and on a circle the same idea returns as arc addition.
Congruent segments are also the defining marks of shapes. A triangle with two ticked sides is isosceles, all four sides of a rhombus are congruent, and the opposite sides of a parallelogram are congruent, which can be proved.
And the slide-and-turn argument is the first case of a bigger idea: two figures are congruent when rigid motions carry one onto the other. For triangles, three pairs of congruent sides are enough, the first of the congruence criteria, and the reason you can build a triangle from three given lengths in only one shape.
Exercises
Reading the notation and the postulate
- Use Figure 1. Name the two segments whose lengths add to AC, and the point they share. Then write the segment addition postulate for Figure 1 as an equation, and rewrite it to give BC.
- Which of these statements make sense, and which do not? (a) \overline{AB} \cong \overline{CD} (b) AB \cong CD (c) AB = CD (d) \overline{AB} = 5
- \overline{JK} \cong \overline{LM} and \overline{LM} \cong \overline{NP}. If JK = 7.4 cm, find LM and NP, and name the property that takes you from \overline{JK} straight to \overline{NP}.
- Five segments have lengths AB = 6 cm, CD = 4.5 cm, EF = 60 mm, GH = 4.5 cm and JK = 5 cm. List every pair of congruent segments.
Finding lengths
- Q is between P and R, PR = 7.3 and QR = 2.8. Find PQ.
- B is between A and C, AB = 2\tfrac{1}{2} and BC = 3\tfrac{3}{4}. Find AC.
- A, B, C and D lie on a line in that order. If AB = 4, BC = 11 and CD = 6, find AD.
- On a grid, one segment runs from (1, 2) to (8, 2) and another from (10, 0) to (10, 7). Find each length and say whether the segments are congruent.
- A triangle has two congruent sides. Its third side is 10 cm and its perimeter is 36 cm. Find the length of each congruent side. Then find the side of a triangle whose three sides are all congruent and whose perimeter is 45 cm, and the side of a square, whose four sides are congruent, with perimeter 34 cm.
- Is the segment from (0, 0) to (6, 8) congruent to the segment from (2, 3) to (12, 3)? Show your working.
Using algebra
- Q is between P and R, PQ = 4x - 7, QR = 2x + 11 and PR = 64. Find x and both pieces.
- B is between A and C, AB = 2x + 1, BC = 3x - 4 and AC = 6x - 12. Find x and all three lengths.
- B is between A and C, AC = 28, and the pieces are in the ratio AB : BC = 2 : 5. Find both pieces.
- M is the midpoint of \overline{AC}, so AM = MC. If AM = 2x + 7 and AC = 6x - 2, find x and AC.
- Two congruent segments have lengths 9 - x and 3x - 7. Find x and the common length.
- Two congruent segments have lengths \displaystyle \frac{2x}{3} + 4 and x - 2. Find x and the common length.
- A triangle has two sides marked with matching ticks, of lengths 2x + 3 and 4x - 9. The third side is x + 5. Find x and the perimeter.
- \overline{PQ} \cong \overline{RS} and \overline{RS} \cong \overline{TU}, with PQ = 5x - 3 and TU = 2x + 9. Find x and RS.
- B is between A and C, AB = x^2 + 1, BC = 5x and AC = 15. Find every value of x that gives a real figure, and the pieces it gives.
- Two congruent segments have lengths x^2 - 9 and 2x + 6. Find every value of x that gives a real segment, and the length it gives.
Several points on a line
Exercises 21 to 25 use four points A, B, C, D on a line in that order, as in Figure 3.
- \overline{AB} \cong \overline{CD}, AD = 20 and BC = 8. Find AB.
- \overline{AB} \cong \overline{CD}, with AB = 3y - 1, CD = y + 7 and BC = 2y. Find y and AD.
- AC = 18, BD = 20 and AD = 29. Find BC, then AB and CD.
- AB = 5, AC = 12 and AD = 30. Find BC, CD and BD.
- \overline{AB} \cong \overline{BC} and \overline{BC} \cong \overline{CD}, so B and C cut \overline{AD} into three congruent pieces, and AD = 27. Find AB and AC.
Which point is between?
For each set of distances, decide whether the three points lie on one line, and if they do, which point is between the other two.
- AB = 6, BC = 9, AC = 12.
- XY = 14, YZ = 5, XZ = 9.
Reasoning
- Explain why, if B is between A and C, the length AB must be less than AC.
- A, B, C and D lie on a line in that order, and AB = CD. Prove, using the segment addition postulate, that AC = BD.
- A, B, C and D lie on a line in that order, and \overline{AB} \cong \overline{CD}. Must \overline{AB} \cong \overline{BC}? Explain with an example, and name a pair of segments that must be congruent.
- B lies on line AC with AB = 7 and BC = 4, but you are not told the order of the points. Explain why the postulate alone cannot give AC, and find every possible value of AC.
- Three points have AB = 4, BC = 7 and AC = 11. Explain why B cannot lie off line AC, and why it cannot lie on line AC outside \overline{AC}.
- Explain why \overline{AB} \cong \overline{BA} for any two points A and B.
- A student measures two segments with a ruler, gets 5.3 cm both times, and writes that the segments are congruent. Is that justified? Explain.
- In the slide-and-turn argument, once A sits on C and the segment lies along ray CD, explain why B must land exactly on D.
Answers
- \overline{AB} and \overline{BC}, which share the point B, between A and C. The postulate is AB + BC = AC; subtracting AB from both sides gives BC = AC - AB.
- (a) and (c) make sense. (b) does not, because AB and CD are numbers and numbers are compared with =. (d) does not, because a segment is a figure, not a number; the intended statement is AB = 5.
- LM = 7.4 cm and NP = 7.4 cm. Congruent segments have equal lengths, so LM = JK, and then NP = LM by the same reason. Going from \overline{JK} to \overline{NP} in one step is the transitive property.
- \overline{AB} \cong \overline{EF} and \overline{CD} \cong \overline{GH}. 60 mm is 60 \div 10 = 6 cm, the same as AB. JK matches nothing.
- PQ = 4.5. The missing piece is the whole minus the known piece: 7.3 - 2.8 = 4.5.
- AC = 6\tfrac{1}{4}. In quarters, \tfrac{10}{4} + \tfrac{15}{4} = \tfrac{25}{4} = 6\tfrac{1}{4}.
- AD = 21. Applying the postulate twice, 4 + 11 + 6 = 21.
- Both have length 7, so they are congruent. The first runs 8 - 1 = 7 squares across and the second 7 - 0 = 7 squares up. One is flat and one upright, but turning does not change a length.
- 13 cm. The two congruent sides together make 36 - 10 = 26, so each is 26 \div 2 = 13. When every side is congruent, the perimeter is shared equally: the triangle’s side is 45 \div 3 = 15 cm and the square’s is 34 \div 4 = 8.5 cm.
- Yes, both have length 10. The first runs 6 across and 8 up, so its length is \displaystyle \sqrt{6^2 + 8^2} = \sqrt{100} = 10. The second is flat and runs 12 - 2 = 10 across.
- x = 10, with PQ = 33 and QR = 31. From (4x - 7) + (2x + 11) = 64, 6x + 4 = 64 and 6x = 60. Check: 33 + 31 = 64.
- x = 9, and the lengths are AB = 19, BC = 23, AC = 42. From (2x + 1) + (3x - 4) = 6x - 12, 5x - 3 = 6x - 12. Check: 19 + 23 = 42.
- AB = 8 and BC = 20. Write the pieces as 2k and 5k. Then 2k + 5k = 28, so 7k = 28 and k = 4.
- x = 8 and AC = 46. M is between A and C with equal pieces, so AM + MC = 2\,AM = AC: 2(2x + 7) = 6x - 2, which is 4x + 14 = 6x - 2. Each piece is 23.
- x = 4, and the common length is 5. From 9 - x = 3x - 7, 16 = 4x. Check: 9 - 4 = 5 and 3(4) - 7 = 5.
- x = 18, and the common length is 16. From \displaystyle \frac{2x}{3} + 4 = x - 2, \displaystyle 6 = \frac{x}{3}. Check: \displaystyle \frac{2(18)}{3} + 4 = 16 and 18 - 2 = 16.
- x = 6, and the perimeter is 41. The ticked sides are equal: 2x + 3 = 4x - 9 gives 2x = 12. The sides are 15, 15 and 6 + 5 = 11, and 15 + 15 + 11 = 41.
- x = 4 and RS = 17. By transitivity \overline{PQ} \cong \overline{TU}, so 5x - 3 = 2x + 9 and x = 4. Then PQ = 17, and \overline{RS} has the same length.
- Only x = 2, giving AB = 5 and BC = 10. The postulate gives x^2 + 1 + 5x = 15, so x^2 + 5x - 14 = 0 and (x + 7)(x - 2) = 0. The root x = -7 would make BC = 5(-7) = -35, and no segment has a negative length, so it is rejected.
- x = 5 only, giving length 16. From x^2 - 9 = 2x + 6, x^2 - 2x - 15 = 0 and (x - 5)(x + 3) = 0. At x = 5 both lengths are 16. At x = -3 both are 0, and a segment’s two endpoints are different points, so its length is never 0; x = -3 is rejected.
- AB = 6. By the segment addition postulate AD = AB + BC + CD, and CD = AB, so 2AB + 8 = 20 and AB = 6.
- y = 4 and AD = 30. From 3y - 1 = y + 7, 2y = 8. Then AB = CD = 11 and BC = 8, so AD = 11 + 8 + 11 = 30.
- BC = 9, AB = 9, CD = 11. Adding the two overlapping segments counts \overline{BC} twice: AC + BD = AD + BC, so BC = 18 + 20 - 29 = 9. Then AB = 18 - 9 and CD = 20 - 9.
- BC = 7, CD = 18, BD = 25. Each is a difference of distances measured from A: 12 - 5 = 7, 30 - 12 = 18 and 30 - 5 = 25.
- AB = 9 and AC = 18. By the segment addition postulate the three pieces add to AD, and congruent pieces have equal lengths, so 3\,AB = 27. Then AC = AB + BC = 9 + 9 = 18.
- Not on one line. The only candidate for the middle is B, the point not touching the longest distance AC = 12, and 6 + 9 = 15, not 12. So no point is between the other two, and the points form a triangle.
- On one line, with Z between X and Y. The longest distance is XY = 14, and the two distances at Z add to it: XZ + ZY = 9 + 5 = 14.
- By the postulate, AC = AB + BC. The length BC is positive, since B and C are different points, so AC is larger than AB.
- B is between A and C, so AC = AB + BC. C is between B and D, so BD = BC + CD. Since AB = CD, the two right sides are equal, so AC = BD. Adding the same piece to two equal pieces gives equal wholes.
- No. The ticks on the end pieces say nothing about the middle one: with AB = CD = 2 and BC = 5 the points are in order and \overline{AB} \cong \overline{CD}, yet AB \ne BC. What must hold is \overline{AC} \cong \overline{BD}, by the shared-piece result.
- AC = 11 or AC = 3. The postulate needs to know which point is in the middle. If B is between A and C, then AC = 7 + 4 = 11. If C is between A and B, then AC + 4 = 7, so AC = 7 - 4 = 3. A cannot be in the middle, since then BA + AC = BC would make the piece BA = 7 longer than the whole BC = 4.
- The distances at B add to the third: AB + BC = 4 + 7 = 11 = AC. If B were off the line, A, B, C would form a triangle, and the triangle inequality would make AB + BC greater than AC. If B were on the line beyond C, then AB = AC + CB = 18, not 4; beyond A, then BC = BA + AC = 15, not 7. So B must be between A and C.
- \overline{AB} and \overline{BA} are the same segment, with the same two endpoints and the points between them. Every segment has the same length as itself, which is the reflexive property.
- No. A ruler is only good to about a millimeter, so the true lengths might differ by a fraction of a millimeter. The readings suggest congruence; a reason from the figure or the given information is needed to prove it.
- The turn keeps B at distance AB from C, and B now lies on ray CD. A ray holds only one point at any given distance from its endpoint, and D is at distance CD = AB. So that one point is D, and B lands on it.