Auxiliary lines
A proof is stuck because the figure is missing the triangle or the parallel line it needs. When may you draw one in, and what may you say about it once it is there?
Where you are 18 in Geometric proof
Why this exists
Some proofs cannot start from the figure they are given. The triangle angle sum is about three angles in three corners, and nothing in a bare triangle puts them side by side. The Base Angles Theorem needs two triangles to compare, and an isosceles triangle is only one. In both, the proof begins by drawing something new: a parallel through the top vertex, a segment that splits the triangle in two.
A line, segment or point added to a figure this way is an auxiliary line, from the Latin for “helping”. It is the most inventive step in geometry, and also the easiest place to cheat, because a line drawn with a pencil can quietly be given properties nothing guarantees. So each added line comes with a reason it exists, and the proof may use only what that reason provides.
Definition
An auxiliary line is a line, ray, segment or point added to a figure during a proof. It is not part of the Given, so it is drawn dashed, and it must be justified: the proof names the fact that guarantees such an object exists.
In Figure 1 the left triangle has a line through A parallel to \overline{BC}. It exists without any postulate, since copying an alternate interior angle makes one, and it is unique by the parallel postulate. The right triangle has \overline{AD} bisecting \angle A, which exists because every angle has exactly one bisector. The matching arcs at A are allowed: the bisector was made that way.
An auxiliary line may be given only the properties its justification supplies. Anything more it turns out to have is a claim, and the proof must prove it.
Intuition
An added line is fixed by two conditions, no more. “Through A” and “parallel to \overline{BC}” pin down one line. “Through A” and “bisects \angle A” pin down one line. A third condition, “and through the midpoint of \overline{BC}”, asks one line to satisfy three demands, and usually none does.
Figure 2 shows it happening. In a triangle with AB \ne AC, the bisector of \angle A meets \overline{BC} at D, and the midpoint of \overline{BC} is M. They are different points, so “the bisector of \angle A through the midpoint of \overline{BC}” is no line at all. A proof that drew it would be using a fact true only of isosceles triangles.
Where an auxiliary line meets the rest of the figure is a position fact, and positions may be read from a figure (what a diagram may assume): that the bisector of \angle A crosses \overline{BC} somewhere between B and C is safe. Exactly where it crosses is a measure fact, and is not.
Results
The justifications. Each move below is guaranteed by a postulate or by an earlier theorem, and each produces exactly one object.
| Added | Why it exists |
|---|---|
| the segment or line joining two points | two points determine exactly one line |
| a segment extended past an endpoint | a segment can be extended as far as needed |
| the midpoint of a segment | every segment has exactly one midpoint |
| the bisector of an angle | every angle has exactly one bisector |
| the perpendicular to a line at a point on it | exactly one such line |
| the perpendicular from a point to a line | exactly one (its length is the distance) |
| the parallel to a line through a point off it | at least one by construction, exactly one by the parallel postulate |
| the point where two lines meet | only once the lines are known to meet |
The rule of two. A new line is named by at most two conditions: two points, or one point and one direction (parallel, perpendicular, or bisecting an angle at that point). Every further property is something to prove.
Why it is true
Adding a line changes nothing about the original figure. The triangle is still the triangle, and the Given still holds. If the added line is guaranteed to exist, every figure that fits the Given can be completed with it, so anything proved about the larger figure, using only the Given and the line’s defining properties, holds for the original.
An unjustified line breaks that chain. It carries an extra assumption into the proof, and the proof then establishes the theorem only for the figures in which that assumption happens to be true, as with the isosceles triangles hidden inside Figure 2’s impossible line. If the extra assumption is part of what was to be proved, the proof is circular.
Euclid built this rule into the Elements. His first two postulates permit joining two points and extending a segment, and before he uses a bisector, a midpoint, a perpendicular or a parallel he constructs it: Book I, Propositions 9 and 10 bisect an angle and a segment, 11 and 12 draw the two perpendiculars, and 31 draws the parallel through a point. From then on each is a legal move, and the same list is what a modern construction has to justify.
Worked examples
Which lines may be drawn?
In a triangle ABC with no two sides equal, say whether each added line is justified. (a) \overline{AM}, where M is the midpoint of \overline{BC}. (b) The line through A parallel to \overline{BC}. (c) The bisector of \angle A, through the midpoint of \overline{BC}. (d) The perpendicular from A to \overleftrightarrow{BC}, meeting it at H.
Solution(a) Justified. \overline{BC} has exactly one midpoint, and two points determine a segment. Two conditions: through A, through M.
(b) Justified. A is not on \overleftrightarrow{BC}, so such a line exists (copy an alternate interior angle), and the parallel postulate makes it the only one. Two conditions: through A, parallel to \overline{BC}.
(c) Not justified. “Through A” and “bisects \angle A” already fix the line, and “through the midpoint” is a third condition. Figure 2 shows it failing when AB \ne AC.
(d) Justified: from a point off a line there is exactly one perpendicular. H is on the line \overleftrightarrow{BC}; if \angle B or \angle C is obtuse, H lies outside the segment (and if one is right, H is that vertex), so the proof must allow for that.
The angles of a quadrilateral
Given. ABCD is a quadrilateral whose two diagonals lie inside it.
Prove. m\angle A + m\angle B + m\angle C + m\angle D = 360^\circ.
Then use it: the angles at A, B, C and D measure (2x + 10)^\circ, (3x)^\circ, (x + 40)^\circ and (4x - 20)^\circ. Find x and each angle.
Draw the diagonal \overline{AC}: two points determine a segment. It lies inside the quadrilateral, so it splits \angle A into \angle 1 and \angle 6 and \angle C into \angle 3 and \angle 4, by the angle addition postulate:
Each triangle’s angles add to 180^\circ:
Adding the two equations, the six angles make 360^\circ, and they regroup into \angle A, \angle B, \angle C and \angle D.
For the numbers, the four expressions add to 360:
The angles are 76^\circ, 99^\circ, 73^\circ and 112^\circ, the ones Figure 3 is drawn with.
Check. 76 + 99 + 73 + 112 = 360.
A bent path, angles on the far side
In Figure 4, m \parallel \ell and the path A, X, B bends at X. The angles between the path and the parts of m and \ell to the left are 140^\circ at A and 125^\circ at B. Find x.
The case with the angles on the near side is in the parallel postulate lesson, where the answer is their sum. Here, first move each given angle to the near side. At A, the 140^\circ angle and the angle between \overline{AX} and m to the right form a linear pair, so that angle is 180^\circ - 140^\circ = 40^\circ. At B it is 180^\circ - 125^\circ = 55^\circ.
Nothing yet connects X to the lines, so add one: through X, the line n \parallel m, which exists because a parallel can be built through any point off a line, is unique by the parallel postulate, and is parallel to \ell as well, since two lines parallel to a third are parallel. With transversal \overline{AX}, the upper part of x and the 40^\circ angle at A are alternate interior angles; with \overline{XB}, the lower part and the 55^\circ angle at B are too. So
In one line: x = 360^\circ - 140^\circ - 125^\circ = 95^\circ. With the far-side angles given, the three angles add to 360^\circ.
A height that must be earned
In \triangle ABC, AB = AC = 13 and BC = 10. Find the distance from A to \overline{BC}.
The distance is measured along the perpendicular, so draw it: from A there is exactly one perpendicular to \overleftrightarrow{BC}. Call its foot D. The square at D is allowed; the perpendicular was made that way.
It is tempting to say BD = 5 because D “is the middle”. That is a second property, and it must be proved. The added line makes two right triangles, \triangle ADB and \triangle ADC, sharing the leg \overline{AD}, so the Pythagorean theorem applies in each:
The right sides are equal because AB = AC, so BD = DC. A point of \overleftrightarrow{BC} at equal distances from B and C can only be the midpoint of \overline{BC}, so D is the midpoint and BD = 5. (Once congruence is available, the same step reads “the triangles are congruent by hypotenuse leg”, a criterion proved later in the course, whose Example 4 is this same figure.)
Now the Pythagorean theorem in \triangle ADB:
The distance is 12, and the triangle’s area is \tfrac{1}{2} \times 10 \times 12 = 60.
Another line, the same answer
In Figure 6, m \parallel \ell, the angle at A is 35^\circ and the angle at B is 50^\circ, both measured to the right. Find x without drawing a parallel through X.
Extend \overline{AX} past X. The line \overleftrightarrow{AX} crosses m at A and is not m, so it crosses \ell as well (a line meeting one of two parallels meets the other); call the crossing P. X lies between the lines, so P lies beyond X.
\overleftrightarrow{AP} is a transversal of m and \ell. The 35^\circ angle at A and \angle XPB are alternate interior angles, so m\angle XPB = 35^\circ.
Now \angle AXB sits outside \triangle XPB, between side \overline{XB} and the extension of \overline{PX} past X: it is an exterior angle. By the exterior angle theorem it equals the two remote interior angles:
That matches the parallel postulate lesson’s answer with a parallel through X. Two different justified lines cannot give different answers, since each proof is a proof about the original figure.
Boundaries
Existence is not free. “Let P be where the lines meet” needs the lines to meet. In Example 5 that took a theorem about parallels; two lines that merely look like they will cross, off the page, guarantee nothing.
Two conditions, not three. An added line may be named by two points, or by a point and a direction. The bisector of \angle A that also passes through the midpoint of \overline{BC}, or the perpendicular from A that also bisects \angle A, exists only in special triangles, and drawing it assumes the triangle is special.
Where it lands is part of the claim. A perpendicular from a vertex can fall outside the opposite side (altitudes), and a diagonal of a concave quadrilateral can lie outside it. When the arrangement can differ, the proof takes each case in turn (proof by cases).
Only in the plane it is justified for. The parallel through a point rests on the parallel postulate, so a proof that adds one works only in Euclidean geometry. The triangle angle sum fails on a sphere precisely because that auxiliary line does not exist there.
Where this goes next
Most auxiliary lines in school geometry do one of two jobs: make a straight angle out of scattered angles, which is how the triangle angle sum is proved, or make two triangles out of one, which is how the Base Angles Theorem is proved. Cutting a polygon into triangles from one vertex gives the interior angle sum of any polygon.
In congruence proofs the added segment is often the shared side of two triangles, which is where planning a congruence proof starts. In circle geometry the usual auxiliary line is a radius, drawn to a point of tangency or to the ends of a chord, and it turns inscribed angles and tangents into facts about isosceles and right triangles.
Exercises
Is the line justified?
In \triangle PQR, no two sides are equal. Say whether each added line is justified, and why.
- The segment from P to the midpoint of \overline{QR}.
- Through Q, the perpendicular to \overleftrightarrow{PR}.
- The bisector of \angle P, meeting \overline{QR} at its midpoint.
- The line through P parallel to \overline{QR}, crossing \overline{PQ} at its midpoint.
- The perpendicular bisector of \overline{QR}, meeting side \overline{PQ} at S.
Cutting polygons into triangles
- Three angles of a quadrilateral measure 95^\circ, 80^\circ and 110^\circ. Find the fourth.
- The angles of a quadrilateral measure x^\circ, (2x)^\circ, (3x)^\circ and (4x)^\circ. Find x and the angles.
- Two diagonals from one vertex cut a pentagon into three triangles. Four of its angles measure 100^\circ, 110^\circ, 120^\circ and 95^\circ. Find the fifth.
- The angles of a quadrilateral measure (3y + 5)^\circ, (2y + 15)^\circ, (4y - 10)^\circ and (y + 50)^\circ. Find y and the angles.
Bent paths between parallels
In Exercises 10 to 12, m \parallel \ell and the path A, X, B bends at X as in Figure 6, with x = m\angle AXB. “Near side” angles are measured to the right, as in Figure 6; “far side” angles to the left, as in Figure 4.
- The near-side angles are 28^\circ at A and 47^\circ at B. Find x.
- The far-side angles are 150^\circ at A and 118^\circ at B. Find x.
- x = 100^\circ, and the near-side angles are y^\circ at A and (3y - 8)^\circ at B. Find y and the angle at B.
- A path runs from A on m to X, then to Y, then to B on \ell, with X and Y between the lines. \overline{AX} runs down to the right and makes 30^\circ with the part of m to the right of A. \angle AXY, opening to the left, is 70^\circ. \overline{YB} runs down to the right and makes 45^\circ with the part of \ell to the left of B. Find m\angle XYB.
Drawing the height
- In \triangle ABC, AB = AC = 17 and BC = 16. Find the distance from A to \overline{BC}.
- An isosceles triangle has equal sides 10 and base 12. Find its height and its area.
- Find the height of an equilateral triangle with side 6.
Reasoning
- A proof of the Base Angles Theorem begins: “Let \overline{AD} be the bisector of \angle A, perpendicular to \overline{BC}.” What is wrong with this first line, and how should it read?
- Given. ABCDE is a convex pentagon. Prove. Its angles add to 540^\circ. Do it by choosing a point O inside the pentagon and joining it to each vertex.
- In quadrilateral ABCD the angle at D is reflex, more than 180^\circ. Explain why the diagonal \overline{AC} does not split it into two triangles, and which diagonal does.
- Prove the result of Example 5 a third way: extend \overline{BX} past X until it meets m at Q.
Answers
- Justified. \overline{QR} has exactly one midpoint, and two points determine a segment.
- Justified. From a point off a line there is exactly one perpendicular. Its foot is on the line \overleftrightarrow{PR}, outside \overline{PR} if \angle P or \angle R is obtuse.
- Not justified. The bisector of \angle P is already fixed, and it passes through the midpoint of \overline{QR} only when PQ = PR.
- Not justified, and impossible. The line through P parallel to \overline{QR} meets \overleftrightarrow{PQ} only at P, which is not the midpoint of \overline{PQ}.
- The line is justified (a midpoint, then the one perpendicular at it), but “meeting side \overline{PQ}” is not. Points on the R side of the bisector are nearer R than Q (the perpendicular bisector theorem, proved later), so P is on the R side exactly when PR < PQ. The bisector therefore crosses \overline{PQ} when PQ > PR, and crosses \overline{PR} instead when PQ < PR. At the boundary PQ = PR it passes through P itself, the case this triangle rules out. The proof must prove which case holds, or take both.
- 75^\circ. A diagonal gives two triangles, so the angles add to 360^\circ, and 360 - 95 - 80 - 110 = 75.
- x = 36; the angles are 36^\circ, 72^\circ, 108^\circ and 144^\circ. From 10x = 360.
- 115^\circ. Three triangles give 3 \times 180^\circ = 540^\circ, and 540 - 100 - 110 - 120 - 95 = 115.
- y = 30; the angles are 95^\circ, 75^\circ, 110^\circ and 80^\circ. The sum is 10y + 60 = 360, so 10y = 300. Check: 95 + 75 + 110 + 80 = 360.
- x = 75^\circ. The parallel through X splits x into 28^\circ and 47^\circ.
- x = 92^\circ. The near-side angles are 180 - 150 = 30 and 180 - 118 = 62, and 30 + 62 = 92. Equivalently 360 - 150 - 118 = 92.
- y = 27 and the angle at B is 73^\circ. From y + (3y - 8) = 100, 4y = 108.
- 85^\circ. Draw parallels through X and through Y. At X the upper part of \angle AXY is 30^\circ (alternate interior with the angle at A), so the lower part is 70^\circ - 30^\circ = 40^\circ. With \overline{XY} as transversal that 40^\circ reappears at Y as the upper part of \angle XYB, and the lower part is 45^\circ (alternate interior with the angle at B). So m\angle XYB = 40^\circ + 45^\circ = 85^\circ.
- 15. The perpendicular from A meets \overleftrightarrow{BC} at D, and as in Example 4, BD^2 = 17^2 - AD^2 = DC^2, so D is the midpoint and BD = 8. Then \displaystyle AD = \sqrt{17^2 - 8^2} = \sqrt{225} = 15.
- Height 8, area 48. The perpendicular from the apex lands at the midpoint of the base (proved as in Example 4), 6 each side, so the height is \displaystyle \sqrt{10^2 - 6^2} = \sqrt{64} = 8 and the area is \tfrac{1}{2} \times 12 \times 8.
- \displaystyle 3\sqrt{3}. An equilateral triangle is isosceles from every vertex, so by the argument of Example 4 the perpendicular from a vertex halves the opposite side, and the height is \displaystyle \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3}, about 5.20.
- It gives \overline{AD} three conditions: through A, bisecting \angle A, and perpendicular to \overline{BC}. The first two fix the line, and the third is a property to prove (it is true in an isosceles triangle, but only because of the theorem being proved). It should read: “Let \overline{AD} bisect \angle A, meeting \overline{BC} at D.”
- Join O to the five vertices: two points determine each segment, and because the pentagon is convex each segment lies inside it. This makes five triangles, whose angles total 5 \times 180^\circ = 900^\circ. The angles at O fill the full turn, 360^\circ. What is left is the pentagon’s five angles, each split in two by a segment: 900^\circ - 360^\circ = 540^\circ.
- With D reflex, \overline{AC} lies outside the quadrilateral, so triangles ABC and ACD overlap instead of fitting together, and the angles at A and C are differences, not sums. The diagonal \overline{BD}, from the reflex vertex, lies inside and splits the quadrilateral into two triangles, so the angle sum is still 360^\circ.
- \overleftrightarrow{BX} meets \ell at B and is not \ell, so it meets m, at a point Q beyond X, to the right of A. It is a transversal of the parallels, so \angle XQA and the 50^\circ angle at B are alternate interior angles: m\angle XQA = 50^\circ. \angle AXB is exterior to \triangle AXQ at X, so x = 35^\circ + 50^\circ = 85^\circ.