The Angle-Bisector Ratio Theorem

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Learning goals

Match opposite-side parts with the adjacent sides in a consistent endpoint order
Prove the internal angle-bisector ratio theorem using an auxiliary parallel
Calculate missing lengths while distinguishing a whole side from its parts
Apply the converse using positive interior side divisions
Distinguish equal angle halves from equal side parts and perpendicular distances

Match each part with its neighboring side

In a nondegenerate triangle $ABC$ , the internal angle bisector from $A$ meets $BC$ at a point $D$ strictly between $B$ and $C$ . It satisfies $\angle BAD\cong\angle DAC$ .

For a numerical picture, take adjacent sides 10 and 15 and an opposite side 20. A drawing of the actual internal angle bisector gives opposite-side parts of about 8 and 12. Their ratio 8:12 matches 10:15, even though the equal angle halves have not cut the opposite side into equal parts.

The angle-bisector ratio theorem states that

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

The part next to $B$ pairs with side $AB$ , and the part next to $C$ pairs with side $AC$ . Reversing both ratios is valid; reversing only one changes the claim.

Equal angle halves, proportional opposite-side parts

Sideways setting Height setting Shape

AD is the internal bisector of angle BAC. D lies strictly between B and C. Angle BAD is approximately 46.6°. Angle DAC is approximately 46.6°. AB is approximately 15.84 units long. AC is approximately 19.73 units long. BD is approximately 11.58 units long. DC is approximately 14.42 units long. BD / DC ≈ 0.802. AB / AC ≈ 0.802. The adjacent sides are unequal, so D is not the midpoint. These are ratios within one triangle, not a dilation scale factor.

Move A while BC stays fixed. Matching arcs show equal half-angles; the two ordered ratios are computed from the drawn lengths and displayed approximately.

Try a symmetric triangle, then move the apex sideways. The displayed decimals are approximate measurements of the drawing. They suggest an exact relationship that the parallel construction below proves.

Worked example 1 Use the part beside B with side AB

Problem. In triangle $ABC$ , $AD$ is the internal bisector of $\angle BAC$ . Suppose

\begin{aligned}AB&=9\text{ cm}\\AC&=15\text{ cm}\\BD&=4.8\text{ cm}.\end{aligned}

Find $DC$ .

BD goes with AB; DC goes with AC. The picture is schematic, so use its labels.

Solution. The two side ratios have the same endpoint order.

\begin{aligned}\frac{4.8}{DC}&=\frac9{15},\\&=\frac35,\\DC&=4.8\cdot\frac53,\\&=8\text{ cm}.\end{aligned}

The longer adjacent side $AC$ goes with the longer opposite-side part $DC$ . The two angle halves are equal even though $4.8$ and $8$ are unequal.

Check your understanding

AD internally bisects angle BAC and $AB:AC=4:7.$ Which ratio must equal 4:7?

BD:DC

DC:BD

BD:BC

AD:BC

A parallel turns equal angles into equal lengths

The two small triangles $ABD$ and $ACD$ are not automatically similar. Equal angles at $A$ give just one corresponding angle pair. Add a parallel to create the relationship we can justify.

Why an internal bisector gives the side ratio#

First consider an actual internal bisector in a triangle with

\begin{aligned}AB&=11\\AC&=17\\BC&=21.\end{aligned}

Through $C$ , construct a parallel to $AD$ that meets $BA$ extended beyond $A$ at $E$ . The parallel angle pairs make $\angle AEC$ and $\angle ACE$ equal to the two angle halves at $A$ . Hence triangle $AEC$ has equal angles at $E,C$ , so

\begin{aligned}AE&=AC,\\&=17\end{aligned}

by the earlier isosceles converse.

In triangle $BEC$ , the parallel cut $AD$ now gives

\begin{aligned}BD:DC&=BA:AE,\\&=11:17.\end{aligned}

The whole base of length $21$ consequently divides into $8.25$ and $12.75$ . We started with the true bisector; we did not choose a point with those lengths and assume it bisected the angle.

Now use general letters. Draw the parallel through $C$ to line $AD$ , meeting line $BA$ at $E$ . The order is $B,A,E$ . To justify it, $B$ and $C$ lie on opposite sides of line $AD$ , since $D$ is interior to $BC$ . The parallel through $C$ cannot cross $AD$ , so its intersection with line $BA$ lies beyond $A$ on the opposite side from $B$ . This reasoning works for acute and obtuse triangles.

The parallel lines give the following two angle equalities.

\begin{aligned}\angle AEC&\cong\angle BAD,\\\angle ACE&\cong\angle CAD.\end{aligned}

The first uses corresponding angles along transversal $BAE$ ; the second uses alternate interior angles along $AC$ . The bisector makes $\angle BAD\cong\angle CAD$ , so the angles at $E$ and $C$ in triangle $AEC$ are equal. Their opposite sides are equal: $AE=AC$ .

The four single-arc angles are equal. Matching side ticks mark AE=AC; B, A, E occur in that order.

In triangle $BEC$ , point $A$ is interior to $BE$ , point $D$ is interior to $BC$ , and $AD\parallel EC$ . Apply the side-splitter theorem, already proved using AA similarity.

\frac{BA}{AE}=\frac{BD}{DC}.

Treat BEC as the whole triangle and AD as its parallel cut. Compare first part with leftover part.

Substitute

\begin{aligned}AE&=AC\\BA&=AB\end{aligned}

to obtain $BD/DC=AB/AC$ , as required.

Check your understanding

In the proof, CE is parallel to AD and B,A,E are in order. Why can AE be replaced by AC?

A Every parallel cut halves a side B The two equal angles at E,C make triangle AEC isosceles C BA and AC are corresponding whole sides D D is always the midpoint of BC

Use the whole side when the parts are unknown

If $AB:AC=m:n$ , then $BD:DC=m:n$ . The whole side $BC$ contains $m+n$ ratio parts. In particular,

\frac{BD}{BC}=\frac{AB}{AB+AC}.

This is a part-to-whole fraction. It is different from $BD/DC=AB/AC$ .

Worked example 2 Divide a known whole side

Problem. The internal bisector $AD$ lies in triangle $ABC$ . Here,

\begin{aligned}AB&=12\\AC&=19\\BC&=24.8.\end{aligned}

Find both $BD$ and $DC$ .

Solution. The side parts have ratio $12:19$ , so the whole has $31$ ratio parts.

\begin{aligned}BD&=24.8\cdot\frac{12}{31},\\&=9.6,\\DC&=24.8\cdot\frac{19}{31},\\&=15.2.\end{aligned}

Their sum is $24.8$ and their ratio is $12/19$ .

The proved part lengths add to the whole base. These are unequal parts despite the equal angle halves.

Worked example 3 Solve an algebraic ratio and check the lengths

Problem. In triangle $ABC$ , $AD$ internally bisects $\angle BAC$ . The lengths are

\begin{aligned}AB&=2x+1\\AC&=3x+4\\BD&=8\\DC&=13.\end{aligned}

Find $x$ and the two adjacent side lengths.

Solution. Keep $B$ first in both ratios.

\begin{aligned}\frac{2x+1}{3x+4}&=\frac8{13},\\13(2x+1)&=8(3x+4),\\26x+13&=24x+32,\\x&=9.5.\end{aligned}

Substitute into the original expressions.

\begin{aligned}AB&=20,\\AC&=32.5.\end{aligned}

Both are positive. Together with $BC=21$ , they satisfy the triangle inequality: the largest length $32.5$ is less than $20+21$ . Also $20/32.5=8/13$ , checking the ratio.

Check your understanding

\begin{aligned}AB:AC&=5:8\\BC&=39\end{aligned}

for an internal bisector AD. What is BD?

24

24.375

15

8

The converse identifies an angle bisector

The converse starts with a nondegenerate triangle and a point $D$ strictly between $B,C$ . If $BD/DC=AB/AC$ , then $AD$ internally bisects $\angle BAC$ .

Why the matching interior ratio identifies the bisector#

For a numerical example, let

\begin{aligned}AB&=18\\AC&=30\\BC&=28.\end{aligned}

Suppose the given interior point has

\begin{aligned}BD&=10.5\\DC&=17.5.\end{aligned}

Construct the actual internal bisector, meeting $BC$ at $D_0$ . The direct theorem makes its parts have ratio $18:30=3:5$ , so they too must be $10.5$ and $17.5$ . Thus $D$ is the true bisector intersection $D_0$ .

A fixed side has only one interior point with this positive division ratio, so the given D and true bisector intersection D₀ coincide.

In general, the actual bisector meets $BC$ at an interior point $D_0$ and the direct theorem gives $BD_0/D_0C=AB/AC$ . Both $D$ and $D_0$ divide the same side with the same positive ratio.

For completeness, if the whole length is $L>0$ and the ratio is $r>0$ , its first part $x$ must satisfy

\begin{aligned}\frac{x}{L-x}&=r,\\x&=r(L-x),\\x(1+r)&=rL,\\x&=\frac{rL}{1+r}.\end{aligned}

This gives one value strictly between $0$ and $L$ . Equal distances from $B$ along segment $BC$ identify the same point, so $D=D_0$ . Therefore the given segment $AD$ is the internal bisector.

Worked example 4 Compare the ratio before claiming equal angles

Problem. Triangle $ABC$ has

\begin{aligned}AB&=14\\AC&=21.\end{aligned}

A point $D$ lies strictly between $B,C$ . Decide whether $AD$ bisects $\angle BAC$ in each case.

Case (a):

\begin{aligned}BD&=6\\DC&=9.\end{aligned}

Case (b):

\begin{aligned}BD&=6\\DC&=10.\end{aligned}

Solution. In case (a), the required ratios agree.

\begin{aligned}\frac69&=\frac{14}{21},\\&=\frac23.\end{aligned}

The converse proves that $AD$ is the internal bisector.

In case (b), $BD/DC=3/5$ , whereas $AB/AC=2/3$ . If $AD$ were an internal bisector, the direct theorem would force equality, so it is not one.

Check your understanding

D is strictly between B,C, with $BD/DC=AB/AC.$ What justifies the conclusion that AD bisects the angle?

A The two small triangles must be congruent B D must be the midpoint C Every segment to BC bisects the angle D The angle-bisector ratio converse

Keep the theorem’s meaning precise

Halving the angle does not usually halve the side. If $AB=AC$ , the theorem gives $BD/DC=1$ , so $D$ is the midpoint. Conversely, if an internal bisector meets the midpoint, then $BD/DC=1$ gives $AB=AC$ . Equal opposite-side parts occur precisely in this isosceles case.

The ratio $AB/AC$ compares two sides of one triangle. It is not an image-to-preimage dilation scale factor, and it does not prove $ABD$ similar to $ACD$ . For example, if $AB\ne AC$ , the proposed matching $A$ to $A$ , $B$ to $C$ , $D$ to $D$ would compare unequal $AB,AC$ but compare shared $AD$ with itself. Those ratios cannot share one scale factor.

This theorem compares lengths along the opposite side with the adjacent side lengths. It does not compare perpendicular distances from a point to the two lines forming the angle. No perpendicular-distance property is needed for this proof or its converse.

Check your understanding

Which lengths does the angle-bisector ratio theorem compare?

A Opposite-side parts and the adjacent side lengths B Perpendicular distances from a point to the angle lines C The bisector length and a circle radius D The two half-angle arc lengths

The Angle-Bisector Ratio Theorem

Learning goals

Match each part with its neighboring side

A parallel turns equal angles into equal lengths

Use the whole side when the parts are unknown

The converse identifies an angle bisector

Keep the theorem’s meaning precise

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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