Triangle angle bisector theorem
A segment cuts one angle of a triangle exactly in half and runs on to the opposite side. Where along that side does it land?
Where you are 10 in Similarity
Why this exists
The bisector of an angle is defined by angles: it cuts the angle into two equal parts. Angle bisectors found what that means for the points on it, which are the points equally far from the two sides. Neither fact says where the bisector of an angle of a triangle meets the side across from that angle.
Symmetry settles one case. When the two sides at the angle are equal, the triangle is its own mirror image across the bisector, and the bisector lands at the midpoint of the opposite side. When the two sides differ, the bisector still lands somewhere, and this theorem says exactly where: it splits the opposite side in the same ratio as the two sides that form the angle. A fact about angles becomes a fact about lengths, which can be computed. It places the meeting point of the three bisectors, the incenter, along each bisector, and it is the reason the bisector and the midpoint of the opposite side part company in every triangle that is not isosceles.
Definition
Let \triangle ABC be a triangle. The bisector from A is the segment that starts at A, cuts \angle BAC into two congruent angles, and ends where it meets \overline{BC}, at a point D. In Figure 1 the two arcs mark the two equal halves.
Triangle angle bisector theorem. If \overline{AD} bisects \angle BAC and D lies on \overline{BC}, then
Read it as: the piece of \overline{BC} next to B, over the piece next to C, equals the side through B, over the side through C. Each piece is paired with the side it touches: \overline{BD} and \overline{AB} both end at B, and \overline{DC} and \overline{AC} both end at C. The third side, \overline{BC}, is the side being cut, and it does not appear on the right.
Intuition
Look at the two small triangles the bisector makes, \triangle ABD and \triangle ADC in Figure 2, and compare their areas in two different ways.
Seen from A. Take \overline{BD} and \overline{DC} as the bases. Both lie on line BC, so both triangles have the same height h, the distance from A to that line. The area of a triangle is half its base times its height, so the areas compare as the bases do: as BD : DC.
Seen from D. Now take \overline{AB} and \overline{AC} as the bases. The height of \triangle ABD is DF, the distance from D to line AB, and the height of \triangle ADC is DG, the distance from D to line AC. Because D lies on the bisector of \angle A, it is the same distance from both sides, so DF = DG. With equal heights, the areas compare as the bases do: as AB : AC.
The same two areas cannot be in two different ratios, so BD : DC = AB : AC. The picture to keep is this: the longer side claims the longer piece. If \overline{AC} is twice as long as \overline{AB}, then \overline{DC} is twice as long as \overline{BD}, and D sits a third of the way from B to C. The bisector leans toward the vertex at the end of the shorter side.
Results
Triangle angle bisector theorem. In \triangle ABC, if the bisector of \angle BAC meets \overline{BC} at D, then
Equivalent form. Multiplying both sides by \displaystyle \frac{DC}{AB} gives each piece over the side it touches:
The two pieces. Write a = BC, b = CA and c = AB. Then
Converse. If D lies strictly between B and C and \displaystyle \frac{BD}{DC} = \frac{AB}{AC}, then \overline{AD} bisects \angle BAC.
The isosceles case. The bisector from A meets \overline{BC} at its midpoint exactly when AB = AC. In every other triangle the bisector from A and the median from A are different segments.
Which piece is longer. BD < DC exactly when AB < AC: the bisector lands nearer the vertex at the end of the shorter side.
Why it is true
The theorem, by a parallel line. This proof turns the bisector into a parallel and then uses the triangle proportionality theorem. Through C, draw the line parallel to \overline{AD}. It is not parallel to line AB, because \overline{AD} is not, so it meets line AB; call the meeting point E. As Figure 3 shows, E lies on the extension of \overline{BA} beyond A.
In \triangle BCE, the segment \overline{AD} joins a point of side \overline{BE} to a point of side \overline{BC} and is parallel to the third side \overline{EC}. The triangle proportionality theorem, with B as the shared vertex, gives
It remains to show that AE = AC. Three angle facts do it.
| Statements | Reasons |
|---|---|
| \angle BAD \cong \angle AEC | corresponding angles, transversal BE across the parallels AD and EC (parallel lines and transversals) |
| \angle DAC \cong \angle ACE | alternate interior angles, transversal AC across the same parallels (alternate angles) |
| \angle BAD \cong \angle DAC | \overline{AD} bisects \angle BAC |
| \angle AEC \cong \angle ACE | the three lines above, chained together |
| AE = AC | in \triangle ACE, sides opposite congruent angles are congruent (isosceles triangles) |
Replace AE by AC in the proportion, and BA is the same length as AB:
The theorem, by areas. The comparison in the Intuition section is already a complete proof. It uses the two ways of computing one area, and the fact that a point on an angle’s bisector is the same distance from both sides, instead of a parallel line.
The two pieces. The theorem says BD : DC = c : b. So \overline{BC}, of length a, is cut into b + c equal shares of \displaystyle \frac{a}{b + c} each, with c of them in \overline{BD} and b of them in \overline{DC}. That gives \displaystyle BD = \frac{ac}{b + c} and \displaystyle DC = \frac{ab}{b + c}.
The converse. The bisector of \angle BAC runs from A into the inside of the angle, and a ray from the vertex into the inside of an angle of a triangle crosses the opposite side (this is called the crossbar theorem). So the bisector meets \overline{BC} at some point D' strictly between B and C, and by the theorem \displaystyle \frac{BD'}{D'C} = \frac{AB}{AC}.
Now slide a point P along \overline{BC} from B to C, with BP = t. Then \displaystyle \frac{BP}{PC} = \frac{t}{a - t}. As t grows the top grows and the bottom shrinks, so the ratio grows the whole way and never takes the same value twice. Only one point between B and C gives the ratio \displaystyle \frac{AB}{AC}. Both D and D' give it, so D = D', and \overline{AD} is the bisector.
The isosceles case and the longer piece. D is the midpoint exactly when \displaystyle \frac{BD}{DC} = 1, which by the theorem is exactly when \displaystyle \frac{AB}{AC} = 1. In the same way, BD < DC means \displaystyle \frac{BD}{DC} < 1, which is \displaystyle \frac{AB}{AC} < 1, that is, AB < AC.
Worked examples
Finding both pieces
In Figure 4, \overline{AD} bisects \angle BAC, with AB = 8, AC = 12 and BC = 15. Find x = BD and y = DC.
The bisected angle is at A, so the sides that form it are \overline{AB} and \overline{AC}. By the theorem the pieces are in the same ratio:
So \overline{BC} is made of 2 + 3 = 5 equal shares, two in \overline{BD} and three in \overline{DC}. Each share is 15 \div 5 = 3, so BD = 6 and DC = 9.
Check. 6 + 9 = 15 and \displaystyle \frac{6}{9} = \frac{2}{3}. The longer piece, \overline{DC}, is the one touching the longer side, \overline{AC}.
A side from the pieces
In \triangle PQR, the bisector of \angle P meets \overline{QR} at S. PQ = 10, QS = 4 and SR = 6. Find PR and the perimeter of the triangle.
SolutionThe bisected angle is at P, so the two sides in the ratio are \overline{PQ} and \overline{PR}. The piece \overline{QS} touches Q, so it pairs with \overline{PQ}; the piece \overline{SR} touches R, so it pairs with \overline{PR}:
Multiply both sides by 6 \cdot PR, which is not zero because PR is a length: 4 \cdot PR = 60, so PR = 15. The third side is QR = 4 + 6 = 10, and the perimeter is 10 + 15 + 10 = 35.
Check. \displaystyle \frac{10}{15} = \frac{2}{3} = \frac{4}{6}. The sides 10, 15 and 10 do make a triangle, since the two shorter ones together, 20, are longer than 15.
An unknown in two sides
In \triangle ABC, the bisector of \angle B meets \overline{AC} at D. AD = 4, DC = 6, AB = x + 3 and BC = 2x. Find x and the two sides.
SolutionThis time the angle is at B. The piece \overline{AD} touches A and pairs with \overline{AB}, and the piece \overline{DC} touches C and pairs with \overline{BC}:
Multiply both sides by 12x, which is not zero because 2x is a length:
So x = 9, which makes AB = 12 and BC = 18.
Check. \displaystyle \frac{12}{18} = \frac{2}{3} = \frac{4}{6}. The third side is AC = 10, and 12 + 10 = 22 > 18, so the three lengths obey the Triangle inequality.
Bisector or median?
In \triangle ABC, AB = 9, AC = 6 and BC = 10. The point D on \overline{BC} has BD = 6, and M is the midpoint of \overline{BC}.
(a) Does \overline{AD} bisect \angle BAC?
(b) How far apart are M and D, and which of them is nearer B?
Solution(a) Showing that a segment is a bisector is the converse’s job, and it needs the two ratios to match. By segment addition DC = 10 - 6 = 4, so
The ratios are equal and D lies strictly between B and C, so by the converse \overline{AD} bisects \angle BAC.
(b) BM = 5 and BD = 6, so MD = 1, and M is the nearer to B. The bisector lands past the midpoint, on the side of C, which is the vertex at the end of the shorter side \overline{AC}. So the median \overline{AM} and the bisector \overline{AD} are different segments, one unit apart where they meet \overline{BC}.
Check. The formula gives the same point: \displaystyle BD = \frac{ac}{b + c} = \frac{10 \cdot 9}{6 + 9} = 6.
All three bisectors
In \triangle ABC, AB = 5, BC = 7 and CA = 8. The bisectors from A, B and C meet the opposite sides at D, E and F. Find all six pieces.
SolutionEach bisector cuts the side across from its own vertex, in the ratio of the two sides at that vertex.
The bisector from A cuts \overline{BC}, in the ratio AB : AC = 5 : 8. That is 13 shares of \displaystyle \frac{7}{13}:
The bisector from B cuts \overline{CA} at E, in the ratio BA : BC = 5 : 7, with the 5 next to A. That is 12 shares of \displaystyle \frac{8}{12} = \frac{2}{3}:
The bisector from C cuts \overline{AB} at F, in the ratio CA : CB = 8 : 7, with the 8 next to A. That is 15 shares of \displaystyle \frac{5}{15} = \frac{1}{3}:
Check. Each pair adds up to its side: \displaystyle \frac{35 + 56}{13} = 7, \displaystyle \frac{10 + 14}{3} = 8 and \displaystyle \frac{8 + 7}{3} = 5. Going round the triangle, the three ratios multiply to 1:
That is no accident. The three bisectors pass through one point, the incenter, and for any three segments from the vertices to the opposite sides that pass through one point, this product is 1. That fact is known as Ceva’s theorem.
On the coordinate plane
The triangle has vertices A = (0, 0), B = (6, 8) and C = (5, 0). The bisector of \angle BAC meets \overline{BC} at D. Find the coordinates of D.
SolutionFirst the two sides at A. By the distance formula, \displaystyle AB = \sqrt{6^2 + 8^2} = 10, and AC = 5 because C is 5 along the x-axis. So BD : DC = 10 : 5 = 2 : 1, and D is two thirds of the way from B to C. To find it, partition the segment: start at B and add two thirds of the step from B to C, which is (5 - 6, 0 - 8) = (-1, -8):
So \displaystyle D = \left(\frac{16}{3}, \frac{8}{3}\right).
Check. A second route gives the direction of the bisector without the theorem. Steps of length 1 from A along the two sides go to \displaystyle \left(\frac{3}{5}, \frac{4}{5}\right) and (1, 0), and those two steps are the sides of a rhombus whose diagonal from A bisects the angle. The diagonal goes to their sum, \displaystyle \left(\frac{8}{5}, \frac{4}{5}\right), so the bisector has slope \displaystyle \frac{4/5}{8/5} = \frac{1}{2}. The segment from A to D has slope \displaystyle \frac{8/3}{16/3} = \frac{1}{2} as well.
Where the incenter sits on a bisector
In \triangle ABC, AB = 6, AC = 9 and BC = 10. The bisector from A meets \overline{BC} at D, and the bisectors from A and B cross at I. Find BD and the ratio AI : ID.
SolutionBy the theorem, BD : DC = AB : AC = 6 : 9 = 2 : 3, so \overline{BC} is 5 shares of 2 and BD = 4.
Now use the theorem a second time, in a smaller triangle. In \triangle ABD, the angle at B is the same angle as \angle ABC, because D lies on \overline{BC}. So the bisector from B bisects the angle of \triangle ABD at B, and it meets the opposite side \overline{AD} at I. The pieces of \overline{AD} are in the ratio of the sides at B:
So AI : ID = 3 : 2, and the incenter is three fifths of the way from A to D.
Check. With the general pieces, \displaystyle \frac{AI}{ID} = \frac{c}{BD} = c \div \frac{ac}{b + c} = \frac{b + c}{a}, and \displaystyle \frac{9 + 6}{10} = \frac{3}{2}.
Boundaries
The ratio uses the sides that form the angle. The two sides in the ratio are the ones meeting at the bisected vertex, and the side being cut never appears in it. Pairing a piece with the wrong side turns \displaystyle \frac{2}{3} into \displaystyle \frac{3}{2}, which puts D on the wrong side of the midpoint.
A bisector is not a median. It reaches the midpoint of the opposite side only in an isosceles triangle with the bisected angle between the equal sides. Example 4 shows the two segments meeting \overline{BC} one unit apart.
Nothing about the bisector’s own length. The theorem fixes where D is, not how long \overline{AD} is. Two triangles can share AB, AC and the ratio BD : DC while the angle at A, and with it \overline{AD}, differs.
The exterior angle has a bisector too. The two lines AB and AC make an angle at A outside the triangle as well, and its bisector behaves like a mirror image of this theorem. When AB \neq AC it meets line BC at a point E outside \overline{BC}, beyond the vertex at the end of the shorter side, and there \displaystyle \frac{EB}{EC} = \frac{AB}{AC} again. In Example 4, with AB = 9 and AC = 6, that point is 20 beyond C, since \displaystyle \frac{30}{20} = \frac{3}{2}. When AB = AC the exterior bisector is parallel to \overline{BC} and meets it nowhere, which Exercise 25 asks you to show.
The converse needs D between B and C. On the whole line BC there are two points with \displaystyle \frac{BD}{DC} = \frac{AB}{AC} (unless the ratio is 1): one between B and C, on the bisector, and one outside, on the exterior bisector. Only the first gives a bisector of \angle BAC.
Where this goes next
Example 7 used the theorem twice to place the incenter on a bisector, at the ratio (b + c) : a from the vertex. Example 5 met Ceva’s theorem, which tests whether three segments from the vertices of a triangle pass through one point, and the angle bisectors are the first family it is usually tried on.
The theorem and its converse are both true and each needs its own argument, a pattern collected in converses and biconditional definitions. The proof by a parallel line is a model for many proofs about similar figures, where the useful line has to be drawn before any ratio appears.
When AB \neq AC, the interior and exterior bisectors at A cut line BC at two points in the same ratio AB : AC. Every point P with \displaystyle \frac{PB}{PC} equal to that ratio lies on one circle, with those two points at the ends of a diameter and A on it as well. That circle, the circle of Apollonius, is where the theorem leads once circles enter the picture.
Exercises
Finding the pieces
In Exercises 1 to 7, the bisector of \angle A in \triangle ABC meets \overline{BC} at D.
- AB = 6, AC = 10 and BC = 12. Find BD and DC.
- AB = 14, AC = 7 and BC = 15. Find BD and DC.
- AB = 9, AC = 12 and BC = 14. Find BD and DC.
- AB = 5, AC = 7 and BC = 9. Find BD and DC.
- AB = 8, BD = 4 and DC = 7. Find AC.
- AC = 15, BD = 6 and DC = 9. Find AB.
- AB = 12, AC = 18 and DC = 9. Find BD and BC.
- In \triangle PQR, the bisector of \angle Q meets \overline{PR} at T. PQ = 20, QR = 16 and PR = 27. Find PT and TR.
Using algebra
The same arrangement as Exercises 1 to 7. Find x and the lengths that depend on it.
- AB = x, AC = 12, BD = 5 and DC = 6.
- AB = 15, AC = 10, BD = x + 1 and DC = x - 1.
- AB = 2x, AC = x + 6, BD = 6 and DC = 5.
- AB = x, AC = 8, BD = 3 and DC = x - 2.
- BD = 4 and DC = 6, and the perimeter of \triangle ABC is 30. Find AB and AC.
Bisector or not
D lies strictly between B and C. Decide whether \overline{AD} bisects \angle BAC, and give the reason.
- AB = 12, AC = 8, BC = 15 and BD = 9.
- AB = 10, AC = 6, BC = 12 and BD = 7. If \overline{AD} is not the bisector, find where the bisector does meet \overline{BC}.
On the coordinate plane
The bisector of \angle BAC meets \overline{BC} at D. Find the coordinates of D.
- A = (0, 0), B = (8, 6) and C = (4, 0).
- A = (0, 0), B = (-6, 8) and C = (15, 0).
Using it elsewhere
- In \triangle ABC, AB = 8, AC = 12 and BC = 10. The bisector from A meets \overline{BC} at D, and the bisectors from A and B cross at I. Find BD and AI : ID.
- A triangular garden has two fences, 36 m and 45 m long, meeting at one corner, and a third fence of 54 m opposite that corner. A straight path leaves the corner, bisecting the angle between the two fences, and runs to the 54 m fence. How far is the end of the path from each end of the 54 m fence?
Reasoning
- In \triangle ABC, AB = AC. Explain why the bisector of \angle A meets \overline{BC} at its midpoint.
- With a = BC, b = CA and c = AB, show from the theorem that \displaystyle BD = \frac{ac}{b + c} and \displaystyle DC = \frac{ab}{b + c}.
- Explain why, if AB < AC, the bisector of \angle A meets \overline{BC} nearer B than C, and closer to B than the midpoint is.
- In \triangle ABC, AB = 4, AC = 6 and BC = 5. Priya says the bisector of \angle A meets \overline{BC} at its midpoint. Explain what is wrong, and find BD.
- Prove the converse: if D lies strictly between B and C and \displaystyle \frac{BD}{DC} = \frac{AB}{AC}, then \overline{AD} bisects \angle BAC.
- In \triangle ABC, AB = AC. Show that the bisector of the exterior angle at A is parallel to \overline{BC}.
Answers
- BD = 4.5 and DC = 7.5. The ratio is 6 : 10 = 3 : 5, so 12 is 8 shares of 1.5.
- BD = 10 and DC = 5. The ratio is 14 : 7 = 2 : 1, so 15 is 3 shares of 5.
- BD = 6 and DC = 8. The ratio is 9 : 12 = 3 : 4, so 14 is 7 shares of 2.
- BD = 3.75 and DC = 5.25. The ratio is 5 : 7, so 9 is 12 shares of 0.75.
- AC = 14. \displaystyle \frac{4}{7} = \frac{8}{AC}, so 4 \cdot AC = 56.
- AB = 10. \displaystyle \frac{6}{9} = \frac{AB}{15}, so \displaystyle AB = \frac{90}{9}.
- BD = 6 and BC = 15. \displaystyle \frac{BD}{9} = \frac{12}{18}, so BD = 6, and BC = 6 + 9.
- PT = 15 and TR = 12. The angle is at Q, so PT pairs with QP and TR with QR: \displaystyle \frac{PT}{TR} = \frac{20}{16} = \frac{5}{4}, and 27 is 9 shares of 3.
- x = 10. \displaystyle \frac{5}{6} = \frac{x}{12}, so 6x = 60.
- x = 5, so BD = 6 and DC = 4. \displaystyle \frac{x + 1}{x - 1} = \frac{15}{10} = \frac{3}{2}, so 2x + 2 = 3x - 3.
- x = 9, so AB = 18 and AC = 15. \displaystyle \frac{2x}{x + 6} = \frac{6}{5}, so 10x = 6x + 36.
- x = 6, so AB = 6 and DC = 4. \displaystyle \frac{x}{8} = \frac{3}{x - 2}, so x(x - 2) = 24, which is (x - 6)(x + 4) = 0. The root x = -4 would make AB negative, so it is rejected.
- AB = 8 and AC = 12. BC = 4 + 6 = 10, so AB + AC = 30 - 10 = 20, and AB : AC = BD : DC = 2 : 3. So 20 is 5 shares of 4.
- Yes. DC = 15 - 9 = 6, and \displaystyle \frac{9}{6} = \frac{3}{2} = \frac{12}{8}. The ratios match and D is between B and C, so the converse applies.
- No. DC = 5, and \displaystyle \frac{7}{5} differs from \displaystyle \frac{10}{6} = \frac{5}{3}, since 7 \cdot 3 = 21 but 5 \cdot 5 = 25. If \overline{AD} were the bisector the theorem would force them to be equal. The bisector meets \overline{BC} at BD = 7.5, since \displaystyle BD = \frac{12 \cdot 10}{10 + 6}.
- \displaystyle D = \left(\frac{36}{7}, \frac{12}{7}\right). \displaystyle AB = \sqrt{8^2 + 6^2} = 10 and AC = 4, so BD : DC = 5 : 2 and D is five sevenths of the way from B to C: \left(8 + \tfrac{5}{7}(-4), 6 + \tfrac{5}{7}(-6)\right).
- \displaystyle D = \left(\frac{12}{5}, \frac{24}{5}\right). \displaystyle AB = \sqrt{6^2 + 8^2} = 10 and AC = 15, so BD : DC = 2 : 3 and D is two fifths of the way from B to C: \left(-6 + \tfrac{2}{5}(21), 8 + \tfrac{2}{5}(-8)\right).
- BD = 4 and AI : ID = 2 : 1. BD : DC = 8 : 12 = 2 : 3, so 10 is 5 shares of 2. In \triangle ABD, the bisector from B meets \overline{AD} at I, so \displaystyle \frac{AI}{ID} = \frac{BA}{BD} = \frac{8}{4}.
- 24 m from the end where the 36 m fence meets it, and 30 m from the other end. The pieces are in the ratio 36 : 45 = 4 : 5, so 54 is 9 shares of 6, and each piece touches the fence it is paired with.
- The theorem gives \displaystyle \frac{BD}{DC} = \frac{AB}{AC} = 1, so BD = DC and D is the midpoint. The bisector from A is then also the median from A.
- The theorem gives BD : DC = c : b, so \overline{BC} is b + c equal shares of \displaystyle \frac{a}{b + c}. \overline{BD} holds c of them and \overline{DC} holds b, so \displaystyle BD = \frac{ac}{b + c} and \displaystyle DC = \frac{ab}{b + c}.
- By the theorem \displaystyle \frac{BD}{DC} = \frac{AB}{AC}, which is less than 1, so BD < DC. Since BD + DC = BC, the smaller piece is less than half of \overline{BC}, so D is closer to B than the midpoint is.
- BD = 2. The bisector lands at the midpoint only when AB = AC. Here BD : DC = 4 : 6 = 2 : 3, so 5 is 5 shares of 1, giving BD = 2 and DC = 3, while the midpoint is 2.5 from B.
- The bisector of \angle BAC crosses \overline{BC} at a point D' strictly between B and C (the crossbar theorem), and by the theorem \displaystyle \frac{BD'}{D'C} = \frac{AB}{AC}. As a point P moves from B to C with BP = t, the ratio \displaystyle \frac{BP}{PC} = \frac{t}{BC - t} keeps growing, so it takes each value only once. D and D' give the same value, so D = D' and \overline{AD} is the bisector.
- Extend \overline{BA} past A to a point P. The exterior angle \angle CAP equals the sum of the two interior angles at B and C, and these are equal because AB = AC (isosceles triangles), so \angle CAP = 2\angle C. Its bisector AQ makes \angle QAC equal to half of that, which is \angle C. Now \angle QAC and \angle ACB are alternate interior angles for the lines AQ and BC with transversal AC, and they are equal, so AQ \parallel BC.