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The Angle Bisector Theorem

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag AA, BB or CC; the foot DD and the perpendiculars follow, and the readout under the canvas keeps BD/DCBD / DC equal to AB/ACAB / AC. Learn more about why it holds
  3. Labels toggles the point names, Measures writes the two angles α\alpha beside their arcs. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?Drop perpendiculars from D to the two sides: DE to AB and DF to AC. The right triangles ADE and ADF share the hypotenuse AD and have equal angles at A, so DE = DF. Then the triangles ABD and ACD have areas ½·AB·DE and ½·AC·DF, in the ratio AB : AC; and since they share the height from A, their areas are also in the ratio BD : DC. Drag any vertex.Angle bisector theorem
Display
Drag A, B or C; D and the feet follow
ααABCD
BD / DC = 0.915, AB / AC = 0.915
Step 1 / 6

Step explanations

1The bisector of the angle at A
Given
AD bisects the angle at A: ∠BAD = ∠DAC = α, and D is the point where the bisector meets BC. To prove: BD : DC = AB : AC.
D is computed from the vertices; the readout shows both ratios. Learn more about the statement

Getting Started with the Visualizer
The Statement
Perpendiculars from D
Two Congruent Right Triangles
The Areas by the Sides
The Areas by the Bases
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag the vertices and watch the foot of the bisector slide along the opposite side so that the two ratios in the readout stay equal. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
1234567
1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes the two angles α beside their arcs
3Draggable points — A, B and C move freely; the hollow foot D of the bisector, and later E, F, H, follow
4Readout — the line under the figure: the ratios BD / DC and AB / AC, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
12345
1The two triangles — ABD and ACD, cut off by the bisector; their areas are compared two ways
2The equal heights — DE = DF, the distances from D to the two sides of the angle
3The shared height AH — dashed, from A to BC: the height of both triangles on the bases BD and DC
4Justification tags — Given, Construction, AAS congruence, Area = ½ base · height (twice), Same ratio
5Replay — at the last step Play becomes Replay and Next is off

The Statement

In triangle ABCABC, let the bisector of the angle at AA meet the opposite side at DD. The theorem, Proposition 3 of Book VI of Euclid's Elements, says that DD divides BCBC in the same ratio as the two sides that form the angle:

BDDC=ABAC\frac{BD}{DC} = \frac{AB}{AC}


The longer side gets the longer piece. If AB=ACAB = AC the bisector is also the median and the altitude, as the isosceles pages show; the theorem says what happens when the two sides differ.

Euclid's proof uses similar triangles. The proof on this page uses areas instead: the two triangles that the bisector cuts off have the same height when measured one way and equal heights when measured another way, and comparing the two measurements gives the ratio. In the player DD and the feet of the perpendiculars are hollow, computed from the vertices. The readout under the canvas shows the two ratios BD/DCBD / DC and AB/ACAB / AC, live while you drag.
ααABCD
Opening scene, frozen

The bisector AD with its two equal angles α, meeting BC at the hollow point D.


Perpendiculars from D

The construction drops two perpendiculars from DD: one to the side ABAB, meeting it at EE, and one to the side ACAC, meeting it at FF. The lengths DEDE and DFDF are the distances from DD to the two sides of the angle, since the perpendicular is the shortest segment from a point to a line.

They will be used as heights. Seen from the base ABAB, the triangle ABDABD has height DEDE; seen from the base ACAC, the triangle ACDACD has height DFDF. The first thing to show is that these two heights are equal.
ααABCDEF
The perpendiculars, frozen

DE to AB and DF to AC, with their right-angle marks.


Two Congruent Right Triangles

Look at the two small right triangles ADEADE and ADFADF, one on each side of the bisector. They share the hypotenuse ADAD. Their angles at AA are both α\alpha, because ADAD bisects the angle. Their angles at EE and FF are both right angles. So two angles agree, and by the angle sum the third angles agree too; with the common side ADAD that makes the triangles congruent (angle-angle-side).

Corresponding sides are equal, so

DE=DFDE = DF


marked with one tick each. This is a theorem in its own right: every point on the bisector of an angle is at the same distance from the two sides of the angle, and it is why the three bisectors of a triangle meet at one point, the centre of the inscribed circle.
ααABCDEF
The congruent right triangles, frozen

ADE and ADF on the common hypotenuse AD: DE = DF, one tick each.


The Areas by the Sides

The bisector cuts the triangle into ABDABD and ACDACD, filled in two colours. Measure their areas with the sides of the angle as bases. For ABDABD, base ABAB and height DEDE; for ACDACD, base ACAC and height DFDF:

[ABD]=12 AB⋅DE,[ACD]=12 AC⋅DF[ABD] = \tfrac{1}{2} \, AB \cdot DE, \qquad [ACD] = \tfrac{1}{2} \, AC \cdot DF


Since DE=DFDE = DF, the heights cancel in the ratio:

[ABD][ACD]=ABAC\frac{[ABD]}{[ACD]} = \frac{AB}{AC}
ααABCDEF
Areas by the sides, frozen

ABD on the base AB with height DE, ACD on the base AC with height DF.


The Areas by the Bases

Now measure the same two areas the other way, with BDBD and DCDC as bases. Both triangles have their apex at AA and their bases on the one line BCBC, so they share the same height: the perpendicular AHAH from AA to BCBC, dashed in the player. Then

[ABD]=12 BD⋅AH,[ACD]=12 DC⋅AH[ABD] = \tfrac{1}{2} \, BD \cdot AH, \qquad [ACD] = \tfrac{1}{2} \, DC \cdot AH


and this time AHAH cancels:

[ABD][ACD]=BDDC\frac{[ABD]}{[ACD]} = \frac{BD}{DC}


Nothing about the bisector was used in this step; it is true for any point DD on BCBC that the two triangles split the area in the ratio of the two pieces of the base.
ααABCDEFH
Areas by the bases, frozen

The same two triangles on the bases BD and DC, sharing the height AH.


Why It Holds

The ratio of the two areas is one number, and it has been written in two ways: as AB:ACAB : AC using the sides of the angle, and as BD:DCBD : DC using the pieces of the opposite side. So

BDDC=ABAC\frac{BD}{DC} = \frac{AB}{AC}


The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given bisector, a construction, angle-angle-side congruence, the area formula twice, and the equality of the two expressions.

The proof shows where each half of the theorem comes from. The step with the bases, [ABD]:[ACD]=BD:DC[ABD] : [ACD] = BD : DC, holds for every point of BCBC; the step with the sides, [ABD]:[ACD]=AB:AC[ABD] : [ACD] = AB : AC, holds exactly when DD is equidistant from the two sides, that is, when DD is on the bisector. So the bisector is the one line through AA for which the two ratios agree, which is also the converse of the theorem.
ααABCDEFH
The conclusion, frozen

The two ways of writing the ratio of the areas agree: BD : DC = AB : AC.


Dragging the Figure

Drag AA sideways: as it moves towards CC the side ACAC shortens, ABAB lengthens, and DD slides towards CC to keep BD/DCBD / DC equal to AB/ACAB / AC; the readout shows both ratios changing together. Put AA straight above the midpoint of BCBC and the two ratios are both 11, the isosceles case. Drag BB or CC to change the base.

The perpendiculars from DD must land on the sides themselves, not on their extensions, for the picture of the two right triangles to make sense, so a drag that makes the angle at BB or at CC obtuse is refused; the theorem itself holds for every triangle. A drag that flattens the triangle or carries a vertex out of the frame is refused too.
ααABCDEFH
After dragging A, frozen

A pulled towards C: AC is now the shorter side and D has slid towards C in the same proportion.


Consequences

• The pieces of the side. With a=BCa = BC, b=ACb = AC, c=ABc = AB, the bisector from AA cuts BCBC into BD=acb+cBD = \dfrac{ac}{b + c} and DC=abb+cDC = \dfrac{ab}{b + c}.
• The incentre. Every point of a bisector is equidistant from the two sides, so the three bisectors of a triangle meet at one point equidistant from all three sides: the centre of the inscribed circle, and the three sides are its tangents, as on the two tangents page.
• The external bisector. The bisector of the exterior angle at AA meets the line BCBC outside the segment, at a point D′D' with BD′:D′C=AB:ACBD' : D'C = AB : AC as well; DD and D′D' divide BCBC internally and externally in the same ratio.
• With trigonometry. The Law of Sines in the two triangles gives BD/sin⁡α=AB/sin⁡∠ADBBD / \sin\alpha = AB / \sin\angle ADB and DC/sin⁡α=AC/sin⁡∠ADCDC / \sin\alpha = AC / \sin\angle ADC, and the two angles at DD are supplementary, so their sines are equal: the theorem in one line.
• Isosceles triangles. When AB=ACAB = AC the ratio is 11 and DD is the midpoint: the bisector, the median and the altitude from the apex are the same line, as on the isosceles base angles page.