# Angle Bisector Theorem: Statement, Proof, Examples

## TL;DR

The angle bisector theorem states that the bisector of an angle in a triangle divides the opposite side into two segments whose ratio equals the ratio of the two adjacent sides: BD/DC = AB/AC. This article covers the statement, the internal and external versions, a clean similar-triangles proof, the converse, the length-of-bisector formula, six worked examples, the common mistakes, and where the theorem leads next.

## A Point on the Opposite Side That Isn't Where You'd Guess

Draw any triangle and bisect one of its angles. The bisector crosses the opposite side at a single point, and that point is not in the middle unless the two neighbouring sides happen to be equal. Its exact position is fixed by a ratio that comes straight from the other two sides of the triangle.

That ratio is the whole theorem, and once you can see why the point sits where it does, every problem built on it becomes a one-line setup.

## What the Angle Bisector Theorem States

In triangle ABC, if the bisector of ∠A meets the opposite side BC at point D, then D divides BC into two segments in the same ratio as the two sides that form ∠A:

\[ \frac{BD}{DC} = \frac{AB}{AC} \].

In words: **the bisector of an angle splits the opposite side into a ratio equal to the ratio of the two adjacent sides.** The segment touching B pairs with the side touching B; the segment touching C pairs with the side touching C. That pairing is the part students most often reverse, so hold on to it.

Using the standard side labels (a = BC, b = CA, c = AB) with m = BD and n = DC, the same statement gives the actual segment lengths:

\[ \frac{m}{n} = \frac{c}{b}, \qquad m = \frac{ac}{b+c}, \qquad n = \frac{ab}{b+c} \].

This is one of the oldest results in geometry: it is Proposition 3 of Book VI of [Euclid's _Elements_](https://mathshistory.st-andrews.ac.uk/Biographies/Euclid/), and it appears in school as [NCERT Class 10, Chapter 6 (Triangles)](https://ncert.nic.in/textbook.php) and under [CCSS-M HSG-SRT.B.5](https://www.thecorestandards.org/Math/Content/HSG/SRT/). The version above is the _internal_ bisector; there is also an external version, which we reach after the proof.

## Why the Theorem Is True — A Proof With Similar Triangles

Before using a result, it helps to see why it holds, and this one needs just a single extra line and a pair of similar triangles.

Start by drawing, through C, a line parallel to the bisector AD, and extend BA until it meets that line at a point E. Two angle facts fall out immediately:

- ∠BAD = ∠AEC, because they are corresponding angles on the parallel lines AD and EC.
- ∠DAC = ∠ACE, because they are alternate interior angles on those same parallels.

Since AD bisects ∠A, the two half-angles ∠BAD and ∠DAC are equal, so ∠AEC = ∠ACE. A triangle with two equal angles is isosceles, which makes triangle ACE isosceles with **AE = AC**.

Now look at triangles BAD and BEC. They share ∠B, and because AD is parallel to EC, they are similar. Similar triangles have proportional sides, so:

\[ \frac{BD}{DC} = \frac{BA}{AE} = \frac{BA}{AC} = \frac{AB}{AC} \].

The parallel line turned the bisector condition into a similar-triangle proportion, and the [isosceles triangle](/content/math/geometry/isosceles-triangles/index.html) linked AE back to AC. (Wikipedia lists four more proofs — by the law of sines, by areas, by altitudes, and by reflection — but this one needs the least machinery, so it is the one worth knowing first.)

## The Converse — A Test for Whether a Line Is Really a Bisector

The proof runs in reverse too, and the reverse statement is genuinely useful. If a line from vertex A meets BC at a point D with

\[ \frac{BD}{DC} = \frac{AB}{AC} \],

then AD must be the bisector of ∠A. So when a problem hands you a cevian (a line from a vertex to the opposite side) and asks whether it bisects the angle, you don't measure the angle at all — you check the two ratios. If the segment ratio matches the side ratio, it bisects; if it doesn't, it doesn't.

## The External Angle Bisector Theorem

Everything so far used the _internal_ bisector, the one that stays inside the triangle. The angle at A also has an _external_ bisector, which splits the angle formed by one side and the extension of the other. It obeys almost the same rule, with one change in where the point lands.

The external bisector of ∠A meets line BC (extended) at a point D′ that divides BC **externally** in the same ratio of the adjacent sides:

\[ \frac{BD'}{CD'} = \frac{AB}{AC} \].

The ratio is identical to the internal case. The difference is the location: D′ sits _outside_ the segment BC, on the extension, so the two distances satisfy \| BD' - CD' \| = BC rather than BD' + CD' = BC. The external version is what lets the theorem keep working even when the relevant point falls outside the triangle, and it is the reason the internal and external bisectors at a vertex are always perpendicular to each other.

## The Length of the Angle Bisector

Knowing where D sits is often enough, but sometimes a problem asks for the actual length of the bisector AD. Once the segment ratio fixes m and n, the length t of the bisector from A follows from applying Stewart's theorem to that cevian, and it simplifies to:

\[ t^2 = bc - mn \].

So the bisector length depends only on the two [adjacent](/content/math/geometry/adjacent/index.html) sides b and c and the two segments m and n it created. We use this directly in the last worked example below.

## A Related Property — and One Theorem It Is Not

One more fact rounds out the picture: **any point on the bisector of an angle is equidistant from the two sides (arms) of that angle.** That property is why the three internal bisectors of a triangle all meet at one point — the incenter — which sits the same distance from all three sides and is the centre of the inscribed circle.

Worth separating clearly: the _angle_ bisector theorem is not the _perpendicular_ bisector theorem. The [perpendicular bisector](/content/math/geometry/perpendicular-bisectors/index.html) of a segment is the set of points equidistant from the segment's two endpoints; the angle bisector theorem is about a ratio of side lengths in a triangle. Some references list them together, but they answer different questions.

## Examples of the Angle Bisector Theorem

With the statement, both types, and the converse in hand, here is the theorem doing actual work. The problems move from a direct ratio split up to a length calculation.

### Example 1: 
In triangle ABC, AB = 8, AC = 12, and the bisector of ∠A meets BC at D. If BC = 10, find BD and DC.

By the theorem, BD/DC = AB/AC = 8/12 = 2/3, so BD : DC = 2 : 3 with BD + DC = 10.

\[ BD = \frac{2}{5} \times 10 = 4, \qquad DC = \frac{3}{5} \times 10 = 6 \].

Final answer: BD = 4, DC = 6.

### Example 2: 
In triangle PQR, PQ = 6, PR = 9, QR = 10. The bisector from P meets QR at S. Find QS.

A first instinct is to pair the segments with the sides by position and write \[ \frac{QS}{SR} = \frac{PR}{PQ} = \frac{9}{6} \], giving QS = (3/5)(10) = 6. Check that against the picture: the bisector from P should land closer to the _shorter_ of the two sides, and PQ = 6 is shorter than PR = 9, so S should sit closer to Q, making QS the _smaller_ segment. A value of 6 out of 10 makes QS the larger one, so the pairing is reversed.

The correct pairing matches each segment to the side sharing its endpoint. Segment QS touches Q, so it pairs with PQ:

\[ \frac{QS}{SR} = \frac{PQ}{PR} = \frac{6}{9} = \frac{2}{3}, \qquad QS = \frac{2}{5} \times 10 = 4 \].

Final answer: QS = 4. In Bhanzu's Grade 10 cohort at the McKinney TX centre, this reversed pairing is the single most common first-attempt error on the theorem — close to five in ten students make it until they start reading the diagram for which side is shorter before writing the ratio.

### Example 3: 
In triangle ABC, the bisector AD meets BC at D with BD = 3, DC = 5, and AB = 6. Find AC.

By the theorem, \( \frac{BD}{DC} = \frac{AB}{AC}, \) so \( \frac{3}{5} = \frac{6}{AC} \). Cross-multiplying gives 3·AC = 30, so AC = 10.

Final answer: AC = 10.

### Example 4: 
In triangle ABC, AB = 2x + 3, AC = 4x − 1, and the bisector from A divides BC so that BD = 5 and DC = 7. Find x.

By the theorem, \( \frac{BD}{DC} = \frac{AB}{AC} \), so \( \frac{5}{7} = \frac{2x + 3}{4x − 1} \). Cross-multiply:

5(4x−1)=7(2x+3); ⇒ 20x−5=14x+21; ⇒ 6x=26; ⇒ x=13/3.

Final answer: x = 13/3 ≈ 4.33.

### Example 5: 
In triangle ABC, AB = 4 and AC = 6. The _external_ bisector of ∠A meets line BC extended at D′. Find the ratio BD′ : CD′.

The external bisector divides BC externally in the ratio of the adjacent sides, exactly as the internal one divides it internally:

\[ \frac{BD'}{CD'} = \frac{AB}{AC} = \frac{4}{6} = \frac{2}{3} \].

Final answer: BD′ : CD′ = 2 : 3, with D′ lying on the extension of BC beyond the closer vertex.

### Example 6: 
In triangle ABC, AB = 5, AC = 7, BC = 6, and AD bisects ∠A. Find the length AD.

First split BC using the theorem, with a = 6, b = AC = 7, c = AB = 5:

\[ m = BD = \frac{ac}{b+c} = \frac{6 \cdot 5}{12} = 2.5, \qquad n = DC = \frac{ab}{b+c} = \frac{6 \cdot 7}{12} = 3.5 \].

Now use the length-of-bisector formula:

\[ t^2 = bc − mn = 35 − (2.5)(3.5) = 35 − 8.75 = 26.25, \qquad t = \sqrt{26.25} \approx 5.12 \].

Final answer: AD ≈ 5.12 units.

## Why the Angle Bisector Theorem Matters

A school theorem earns its place by what it unlocks downstream, and this one unlocks a lot.

- **The incenter.** Because every point on an angle bisector is equidistant from the angle's two sides, the three bisectors of a triangle meet at one point that is equidistant from all three sides. That point is the incenter, the centre of the largest circle that fits inside the triangle — used anywhere a circle must be inscribed in a triangular boundary.

- **Lengths without coordinates.** The theorem turns a length problem into a ratio, and the ratio into one line of [algebra](/content/math/algebra/index.html). Paired with the length-of-bisector formula (which comes from Stewart's theorem), it handles cevian problems that would otherwise need full coordinate geometry.

- **Olympiad and JEE geometry.** Spotting an angle bisector inside a configuration and applying the ratio is a standard first move in competition problems, and the external version regularly appears in JEE Advanced questions about points lying on extended sides.

- **Real aim-and-bounce problems.** The equidistant-from-both-sides property is the [geometry](/content/math/geometry/index.html) behind reflection paths — the angle a ball takes off a billiard cushion, or the way a signal reflects at equal angles — which is why the theorem turns up in physics and design as well as in pure geometry.

## Tripping Points to Avoid With Angle Bisector Theorem

### Mistake 1: Pairing the segment with the wrong side

**Where it slips in:** A student writes BD/DC = AC/AB, matching segments to sides by their position in the figure instead of by shared vertex.

**Don't do this:** Pair BD with AC just because they look opposite on the page.

**The correct way:** Match by shared endpoint. Segment BD touches B, so it pairs with the side touching B, which is AB. The bisector always lands closer to the shorter side, so the smaller segment sits next to the shorter side — a quick sanity check before you commit to a ratio.

### Mistake 2: Using the theorem on a cevian that isn't a bisector

**Where it slips in:** A figure shows a line from a vertex that looks symmetric, and the student applies the ratio without checking that it actually bisects the angle.

**Don't do this:** Trust a diagram that merely looks balanced.

**The correct way:** Use the converse. Compute the segment ratio and the side ratio; only if they match is the line a bisector and the theorem valid. If they differ, reach for Stewart's theorem or [similar triangles](/content/math/geometry/similar-triangles/index.html) instead.

### Mistake 3: Mixing up internal and external division

**Where it slips in:** An external-bisector problem is solved as if the point lay inside BC, so the student writes BD′ + CD′ = BC.

**Don't do this:** Assume the same ratio always gives a point between B and C.

**The correct way:** Internal bisector means D lies inside BC, so BD + DC = BC. External bisector means D′ lies outside BC on the extension, so the distances satisfy \|BD' − CD'\| = BC. The ratio is the same; the placement is not.

## Key Takeaways

- The **angle bisector theorem** says a triangle's angle bisector divides the opposite side in the ratio of the two adjacent sides: BD/DC = AB/AC.
- Pair each segment with the side that shares its vertex, and remember the bisector lands closer to the shorter side.
- The internal bisector divides BC inside the segment; the external bisector divides it externally in the same ratio, outside the segment.
- The converse — equal ratios imply a bisector — is the test for whether any cevian actually bisects the angle.
- The bisector's length is t² = bc − mn, and the three internal bisectors meet at the incenter.