Constructing a triangle from given parts
You are handed three lengths, or two lengths and an angle, or two angles and a length. Is there a triangle with exactly those parts, how do you draw it with only a compass and a straightedge, and is the one you draw the only one?
Where you are 8 in Compass and straightedge constructions
When you reach for it
A triangle has six parts: three sides and three angles. The SSS, SAS, ASA, AAS say that some sets of three parts already decide all six: three sides (SSS), two sides and the angle between them (SAS), two angles and the side between them (ASA). This lesson is the other half of that statement. A criterion says two triangles with those parts must match. A construction shows that a triangle with those parts exists, and draws it.
Reach for it whenever a figure has to be built from given parts: laying out a triangle whose sides are given lengths, deciding whether three lengths can be the sides of a triangle at all, or starting a larger construction whose first move is a triangle with chosen sides.
The tools are those of every construction: an unmarked straightedge and a compass that holds its opening. The given parts arrive as drawn segments and drawn angles. The two moves used throughout, setting the compass to a segment and swinging it from a new point, and copying an angle onto a ray, are taught in copying a segment and an angle. Numbers such as 6 cm appear in the examples only so the results can be checked; no step reads a scale. With a ruler and protractor instead, the same questions are the subject of Drawing triangles from given sides and angles: one, many or none.
In \triangle ABC the side opposite each vertex is named by the matching small letter, so a = BC, b = CA and c = AB.
The procedure
Three sides
You are given three segments, of lengths a, b and c. The goal is \triangle ABC with BC = a, CA = b and AB = c.
Lay down one side. Draw a ray from a point A and copy the segment of length c onto it, ending at B, so AB = c. Any of the three lengths can go first; taking the longest keeps the triangle sitting above it.
An arc for each of the other sides. Set the compass to b and draw an arc about A, above the line and below it. Set the compass to a and draw an arc about B in the same way. Call a point where the two arcs cross C, and draw \overline{AC} and \overline{BC}.
Why the sides are right. C lies on the circle with center A and radius b, and a circle is the set of points at that one distance from its center, so CA = b. It also lies on the circle with center B and radius a, so CB = a. The side AB = c was laid down first. The angles were never looked at; they come out however the three lengths force them.
Why it is the only one. The arcs cross twice, at C above \overline{AB} and at C' below it. \triangle ABC and \triangle ABC' look different on the page, but they have the same three side lengths, so they are congruent by SSS; one is the reflection of the other in line AB. More than that, any triangle at all with sides a, b and c is congruent to the one drawn, by the same criterion. So the construction does not find a triangle with these sides; it finds the triangle, and only its position on the page was a choice.
When the arcs meet. Two circles whose centers are a distance d apart, with radii r_1 and r_2, cross at two points exactly when
Here the centers are A and B, so d = c, and the radii are b and a. The right-hand inequality says c < a + b, and the left-hand one says both b < a + c and a < b + c. Together they are the Triangle inequality: each side is shorter than the sum of the other two. If c > a + b, the arcs stop short of each other along \overline{AB} and never meet. If c = a + b exactly, they touch at a single point on \overline{AB} itself, and the figure is flat.
Only one of the three inequalities ever needs checking: the longest length against the sum of the other two. The other two inequalities each say that a length is less than the longest plus something positive, which is always true.
Two sides and the angle between them
You are given two segments, of lengths b and c, and an angle. The goal is \triangle ABC with AB = c, AC = b and \angle A congruent to the given angle.
Copy the angle. Copy the given angle at a point A, so that it has two arms from A.
Mark the two sides along the arms. Set the compass to c and swing an arc from A across one arm; call the crossing B. Set it to b and swing an arc from A across the other arm; call the crossing C. Draw \overline{BC}.
Why it works. Each arm is a ray from the center of the arc, so it meets the arc exactly once. B and C are determined, and AB = c, AC = b by the definition of a circle. The angle at A is a copy of the given one. Any other triangle with these two sides and this angle between them is congruent to \triangle ABC by SAS. The third side and the other two angles are not chosen; they are whatever they come out to be.
No condition on the lengths. Any two lengths and any angle strictly between 0^\circ and 180^\circ give a triangle. The two arms leave A in different directions that are not opposite, so A, B and C never lie on one line.
Two angles and the side between them
You are given a segment of length c and two angles, \angle 1 and \angle 2. The goal is \triangle ABC with AB = c, \angle A \cong \angle 1 and \angle B \cong \angle 2.
Lay down the side, then copy an angle at each end. Copy the segment as \overline{AB}. Copy \angle 1 at A with one arm along \overrightarrow{AB}, and \angle 2 at B with one arm along \overrightarrow{BA}, both opening on the same side of line AB. Call the point where the two new arms cross C.
Why it works. The angles at A and B are copies of the given ones, the side between them is a copy of the given side, and any other triangle with these three parts is congruent to \triangle ABC by ASA.
When the arms meet. The new arms cross exactly when m\angle 1 + m\angle 2 < 180^\circ. If the sum is 180^\circ, the arms are parallel: they make same-side interior angles with line AB that are supplementary. If the sum is more, the arms spread apart. That a sum below 180^\circ is enough is Euclid’s fifth postulate, which says that when a line crossing two lines makes interior angles on one side that together are less than two right angles, the two lines meet on that side; see The parallel postulate.
Two angles and a side that is not between them. If the given side is opposite one of the given angles (AAS), find the third angle first. By the Triangle angle sum it is 180^\circ minus the other two, and it can be made without measuring by copying the two given angles side by side on a line and keeping what is left of the straight angle, as in copying a segment and an angle. The given side now lies between two known angles, and the construction above applies.
Worked examples
Three given sides
Construct a triangle with sides 6 cm, 4 cm and 5 cm, and say how many different triangles there are.
SolutionFirst check that the arcs will meet. The longest length is 6, and the other two add to 4 + 5 = 9, which is more than 6. So the triangle inequality holds, and the other two inequalities hold with it.
Copy the 6 cm segment as \overline{AB}. Set the compass to 4 cm and draw an arc about A; set it to 5 cm and draw an arc about B. The arcs cross at C above the line and at C' below it. Then AC = 4 cm, because C is on the circle about A of radius 4 cm, and BC = 5 cm for the same reason about B. This is the triangle in the first figure.
There is one triangle. \triangle ABC and \triangle ABC' are reflections of each other in line AB, and any triangle with sides 6, 4 and 5 is congruent to both by SSS.
Which three lengths make a triangle
For each set of lengths, decide whether the three-sides construction produces a triangle: (a) 3, 4, 8; (b) 5, 5, 10; (c) 7, 9, 12; (d) 2, 9, 10.
SolutionIn each case lay down the longest length as \overline{AB} and draw the arcs of the other two about A and B. Only the longest needs comparing with the sum of the other two.
(a) 3 + 4 = 7, less than 8. The arc of radius 3 about A reaches only 3 along \overline{AB}, the arc of radius 4 about B reaches 4 back, and a gap of 8 - 7 = 1 is left between them. Every other point of the arc about A is further from B than that nearest point, so the arcs never meet. No triangle.
(b) 5 + 5 = 10, equal to the longest. Both arcs reach the midpoint of \overline{AB} and touch there, and nowhere else. The only candidate for C lies on \overline{AB}, so A, B and C are on one line. No triangle; the figure collapses to a segment.
(c) 7 + 9 = 16, more than 12. The arcs cross twice, and there is one triangle, up to reflection.
(d) 2 + 9 = 11, more than 10. A triangle, even though one side is very short. The other two sides differ by only 10 - 9 = 1, which is less than 2, so the short side can still close the gap between them.
The possible third side
Two sides of a triangle are 7 cm and 4 cm. What lengths can the third side have? How many of them are whole numbers of centimeters?
SolutionLay down the 7 cm side as \overline{AB} and draw the circle of radius 4 cm about B. The third vertex C must be on that circle, and the third side is AC. As C moves round the circle, AC is smallest where the circle crosses \overline{AB}, at 7 - 4 = 3 cm from A, and largest on the far side of B, at 7 + 4 = 11 cm from A. Those two points lie on line AB, where the triangle is flat, so they are left out. Every length strictly between them is reached by a point of the circle off the line. So the third side x satisfies
The triangle inequality says the same: x < 7 + 4, and 7 < 4 + x gives x > 3. The third inequality, 4 < 7 + x, holds for every positive x.
The whole numbers strictly between 3 and 11 are 4, 5, \ldots, 10, which is 10 - 4 + 1 = 7 lengths.
Two sides and a right angle between them
A right angle is drawn on the page. Construct a triangle with sides of 6 cm and 8 cm along its arms, and find the length of the third side.
SolutionCopy the right angle at A. Swing an arc of radius 6 cm from A across one arm, giving B, and an arc of radius 8 cm across the other, giving C. Draw \overline{BC}. By SAS this is the only triangle with these two sides and a right angle between them.
The third side was not chosen, but it is fixed. \triangle ABC has its right angle at A, so by the The Pythagorean theorem
This is what SAS means in practice: two sides and the angle between them leave nothing undecided, so the third side can be computed as well as drawn. For an angle other than a right angle, the computation is the law of cosines.
Two angles and a side, in either position
(a) Construct a triangle with a side of 7 cm and angles of 40^\circ and 65^\circ at its ends, and find its third angle. (b) Construct \triangle ABC with m\angle A = 40^\circ, m\angle B = 65^\circ and BC = 7 cm.
Solution(a) Copy the 7 cm segment as \overline{AB}. Copy the 40^\circ angle at A and the 65^\circ angle at B, on the same side of line AB. Their sum, 105^\circ, is less than 180^\circ, so the new arms meet at a point C, as in the third figure. By the triangle angle sum,
(b) Now the 7 cm side, \overline{BC}, runs from the 65^\circ angle at B to the vertex C, whose angle is not given; it is opposite the 40^\circ angle, not between the two given ones. The third angle is again 75^\circ by the same sum, and it can be made without measuring by copying the 40^\circ and 65^\circ angles side by side on a line and keeping what is left. Copy the 7 cm segment as \overline{BC}, copy the 65^\circ angle at B and the 75^\circ angle at C on the same side, and let the arms meet at A. The angle there is 180^\circ - 65^\circ - 75^\circ = 40^\circ, as required.
The two triangles have the same three angles but are not congruent. In (a) the 7 cm side lies opposite the 75^\circ angle; in (b) it lies opposite the 40^\circ angle. Which angle a length sits opposite matters as much as the length itself.
Two sides and an angle that is not between them
m\angle A = 30^\circ, AB = 8 cm and BC = 5 cm. Try to construct \triangle ABC, and find every possible length of \overline{AC}.
SolutionCopy the 30^\circ angle at A and mark B on one arm with AB = 8 cm. The vertex C must lie on the other arm and be 5 cm from B, so swing an arc of radius 5 cm about B and see where it meets that arm.
It meets it twice. To find where, drop the perpendicular from B to the arm, meeting it at H. In the right triangle \triangle AHB the angle at A is 30^\circ, so the side opposite it is half the hypotenuse (the 30^\circ-60^\circ-90^\circ triangle): BH = 4. By the Pythagorean theorem,
Both crossings are 5 cm from B, so in the right triangles with legs along \overline{BH} and the arm,
Since \displaystyle 3 < 4\sqrt{3}, both crossings are on the arm itself, one on each side of H, and
about 9.93 cm and 3.93 cm. \triangle ABC and \triangle ABD both have a 30^\circ angle at A, a side of 8 cm from A to B and a side of 5 cm from B, yet they are not congruent: one third side is more than twice the other. This is why two sides and an angle not between them is not a congruence criterion, as Why AAA and SSA fail explains. How many triangles there are depends on how BC compares with the height BH and with AB, which is the height test: shorter than BH, none; equal to BH, one right triangle; between BH and AB, two; at least AB, one.
Boundaries
Three lengths need the triangle inequality, strictly. The longest must be less than the sum of the other two. Equal gives a flat figure on one line; more gives arcs that never meet.
Two sides and the angle between them always work, for any two lengths and any angle strictly between 0^\circ and 180^\circ.
Two angles need a sum below 180^\circ. At exactly 180^\circ the arms are parallel, and above it they spread apart. When the sum is below, the third angle follows from the angle sum.
Two sides and an angle not between them can give none, one or two triangles. The construction draws every one of them. It cannot pick between two, because the given parts do not pick.
Three angles fix no triangle. There is no length to lay down, and triangles of every size share the same three angles. Angles alone decide the shape at most, never the size.
Each triangle appears twice on the page. The arcs or arms can be drawn on either side of the first side, and the two results are mirror images, congruent to each other. “The only triangle” always means the only one up to congruence.
Only the two tools. The numbers in the examples are there to check results; the construction itself never reads a scale or a protractor.
Where this goes next
The three-sides construction with all three lengths equal is the equilateral triangle, drawn with a single compass setting; it is the first proposition of Euclid’s Elements, and the general three-sides case is Book I, Proposition 22. Every construction defended by congruent triangles, as in why a construction works, relies on what is shown here: the parts in a congruence criterion really do produce a triangle.
Computing the parts a construction leaves undecided, the third side in SAS or the angles in SSS, is the work of the law of cosines and the law of sines, and the two triangles of Example 6 return there as the ambiguous case. The circle through the three vertices of a constructed triangle is found in constructions with circles.
Exercises
Three sides
- Decide whether each set of lengths gives a triangle by the three-sides construction: (a) 4, 7, 10; (b) 3, 5, 9; (c) 6, 6, 12; (d) 2.5, 6, 6.
- Two sides of a triangle are 9 cm and 5 cm. What lengths can the third side have, and how many of them are whole numbers of centimeters?
- Two sides of a triangle are both 6 cm. What lengths can the third side have, and how many of them are whole numbers of centimeters?
- \overline{AB} is 10 cm long, and an arc of radius 5 cm is drawn about B. For which radii r does an arc about A cross it at two points?
- On a coordinate grid, A = (0, 0) and B = (8, 0). Arcs of radius 5 are drawn about A and about B. Where do they cross?
- On a coordinate grid, A = (0, 0) and B = (14, 0). An arc of radius 13 is drawn about A and an arc of radius 15 about B. Where do they cross?
- A triangle is to have a 12 cm side, and its other two sides together are to make 20 cm, one of them x cm. For which x is there a triangle?
Two sides and the angle between them
- A triangle has sides of 5 cm and 12 cm with a right angle between them. How long is the third side?
- A triangle has two sides of 7 cm with a 60^\circ angle between them. Find its other two angles and its third side.
- A triangle has two sides of 4 cm with a 120^\circ angle between them. Find its other two angles.
Two angles and a side
- Angles of 50^\circ and 75^\circ are copied at the two ends of a 6 cm segment, on the same side. What angle do the arms make where they meet?
- Angles are copied at the two ends of a segment, on the same side. Do the new arms meet when the angles are (a) 95^\circ and 85^\circ, (b) 100^\circ and 90^\circ, (c) 20^\circ and 150^\circ? Where they meet, give the third angle.
- \triangle ABC is to have m\angle A = 35^\circ, m\angle B = 80^\circ and BC = 6 cm. The 6 cm segment is copied as \overline{BC}. Which angles should be copied at its two ends?
Two sides and an angle not between them
- m\angle A = 30^\circ and AB = 10 cm, and an arc of radius BC about B is swung across the other arm of \angle A. How many triangles are there when BC is (a) 4 cm, (b) 5 cm, (c) 7 cm, (d) 12 cm?
- In Exercise 14(b), find AC.
- In Exercise 14(c), find both possible lengths of AC.
Reasoning
- Explain why, for three lengths, it is enough to check that the longest is less than the sum of the other two.
- In the three-sides construction the arcs cross at C and C'. Prove that \triangle ABC \cong \triangle ABC', and explain why C' is the reflection of C in line AB.
- Explain why two sides and the angle between them always give a triangle, whatever the lengths.
- Explain why three angles, such as 50^\circ, 60^\circ and 70^\circ, cannot be turned into a construction that produces one triangle.
Answers
- (a) Yes: 4 + 7 = 11 > 10. (b) No: 3 + 5 = 8 < 9, so the arcs stop short of each other. (c) No: 6 + 6 = 12, so the arcs touch on the segment and the figure is flat. (d) Yes: 2.5 + 6 = 8.5 > 6, and the two 6 cm sides make it isosceles.
- 4 < x < 14, and 9 whole lengths. The third side is more than 9 - 5 = 4 and less than 9 + 5 = 14; the whole numbers 5 to 13 number 13 - 5 + 1 = 9.
- 0 < x < 12, and 11 whole lengths. The two sides differ by 6 - 6 = 0, so any positive length below 6 + 6 = 12 works; the whole numbers 1 to 11 number 11 - 1 + 1 = 11.
- 5 < r < 15. Two crossings need |r - 5| < 10 < r + 5. The right-hand inequality gives r > 5; the left-hand one gives r - 5 < 10, so r < 15, and 5 - r < 10 holds for every positive r.
- (4, 3) and (4, -3). A crossing (x, y) has x^2 + y^2 = 25 and (x - 8)^2 + y^2 = 25. Subtracting gives 16x - 64 = 0, so x = 4, and then y^2 = 9.
- (5, 12) and (5, -12). Subtracting (x - 14)^2 + y^2 = 225 from x^2 + y^2 = 169 gives 28x - 196 = -56, so x = 5, and then y^2 = 169 - 25 = 144.
- 4 < x < 16. The other side is 20 - x. The triangle inequality needs x + 12 > 20 - x, so x > 4, and (20 - x) + 12 > x, so x < 16. The third inequality, x + (20 - x) = 20 > 12, always holds.
- 13 cm. By the Pythagorean theorem, \displaystyle \sqrt{5^2 + 12^2} = \sqrt{169} = 13.
- Both 60^\circ, and the third side is 7 cm. The triangle is isosceles, so the angles opposite the equal sides are equal (Proving the base angles theorem), each (180^\circ - 60^\circ) \div 2 = 60^\circ. All three angles are equal, so by the converse of the base angles theorem all three sides are equal too.
- Both 30^\circ. The triangle is isosceles, so the other two angles are equal, each (180^\circ - 120^\circ) \div 2 = 30^\circ.
- 55^\circ. The sum 50^\circ + 75^\circ = 125^\circ is below 180^\circ, so the arms meet, and the third angle is 180^\circ - 50^\circ - 75^\circ = 55^\circ.
- (a) No: 95^\circ + 85^\circ = 180^\circ, so the arms are parallel. (b) No: 100^\circ + 90^\circ = 190^\circ, so they spread apart. (c) Yes: 20^\circ + 150^\circ = 170^\circ, and the third angle is 10^\circ.
- 80^\circ at B and 65^\circ at C. The ends of \overline{BC} are B and C, so the angles there are \angle B and \angle C, and m\angle C = 180^\circ - 35^\circ - 80^\circ = 65^\circ.
- The height from B to the arm is half of AB, 10 \div 2 = 5 cm, by the 30^\circ-60^\circ-90^\circ triangle. (a) None, since 4 < 5. (b) One, a right triangle with the right angle at C. (c) Two, since 5 < 7 < 10. (d) One, since 12 > 10: the second crossing falls on the ray opposite the arm, behind A.
- \displaystyle AC = 5\sqrt{3} cm, about 8.66 cm. C is the foot of the perpendicular from B, so \displaystyle AC = \sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt{3}.
- \displaystyle 5\sqrt{3} + 2\sqrt{6} cm and \displaystyle 5\sqrt{3} - 2\sqrt{6} cm, about 13.56 cm and 3.76 cm. The foot H of the perpendicular is \displaystyle 5\sqrt{3} from A, and each crossing is \displaystyle \sqrt{7^2 - 5^2} = \sqrt{24} = 2\sqrt{6} from H, one on each side.
- Call the lengths a \le b \le c. The three inequalities are c < a + b, b < a + c and a < b + c. Since c \ge b and a > 0, a + c > b always; since c \ge a and b > 0, b + c > a always. Only c < a + b can fail.
- AC = AC', since both points are on the circle of radius b about A; BC = BC' likewise, on the circle of radius a about B; and \overline{AB} is shared. So \triangle ABC \cong \triangle ABC' by SSS. Since A is equally far from C and C', it lies on the perpendicular bisector of \overline{CC'}, and so does B. Two different points fix a line, so that bisector is line AB. A line that is the perpendicular bisector of \overline{CC'} is exactly the mirror that reflects C to C'.
- The angle lies strictly between 0^\circ and 180^\circ, so its two arms leave A in directions that are neither the same nor opposite. An arc about A of any radius meets each arm exactly once, because each arm is a ray from the center. So B and C always exist, and they cannot lie on one line with A; three points not on one line make a triangle. No inequality between the lengths is ever needed.
- Each step of a construction places a point at a given distance, or along a given direction, from points already drawn, and three angles supply no distance. Starting from any segment and copying two of the angles at its ends gives a triangle by ASA, but a longer starting segment gives a larger triangle with the same three angles. Congruent triangles have equal sides, so these triangles are not congruent, and the angles alone cannot pick one of them. They fix the shape, not the size.