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The Angle-Bisector Ratio Theorem: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Match the side next to each endpoint

In triangle $ABC$ , $AD$ is the internal bisector of $\angle BAC$ , with $D$ strictly between $B,C$ . The lengths are
$\begin{aligned}AB&=8.4\text{ cm}\\AC&=13.2\text{ cm}\\BD&=7\text{ cm}.\end{aligned}$
Find $DC$ and the whole length $BC$ . Explain the endpoint order used in your proportion.

Pair each base part with the adjacent side touching its endpoint.

Text description of this figure

AD is a marked internal bisector. The schematic labels AB8.4, AC13.2 and BD7; DC is unknown and no calculated length is shown.
1. Hint 1
  The part beside B pairs with the adjacent side ending at B.
2. Hint 2
  Find the leftover base part first; the whole base is the sum of its two parts.
Answer

$\begin{aligned}DC&=11\text{ cm}\\BC&=18\text{ cm}.\end{aligned}$

Full solution

The theorem pairs $BD$ with $AB$ and $DC$ with $AC$ .
$\begin{aligned}\frac 7{DC}&=\frac{8.4}{13.2},\\&=\frac 7{11},\\DC&=11\text{ cm}.\end{aligned}$

Add the two base parts:
$\begin{aligned}BC&=7+11,\\&=18\text{ cm}.\end{aligned}$
Both the parts and the adjacent sides have ratio $7:11$ .

Answer

$\begin{aligned}DC&=11\text{ cm}\\BC&=18\text{ cm}.\end{aligned}$

Key idea

Keep the same neighboring endpoint first in both ratios.
Problem 2 A known base contains both ratio parts

An internal angle bisector $AD$ lies in triangle $ABC$ . Here,
$\begin{aligned}AB&=17\\AC&=23\\BC&=30.\end{aligned}$
Find $BD$ and $DC$ . Give both a sum check and a ratio check.

The base is a whole, not the leftover part.

Text description of this figure

The schematic shows the internal bisector and whole base 30, with adjacent sides 17 and 23. Both opposite-side parts are unknown.
1. Hint 1
  The whole base contains the sum of the two ratio counts.
2. Hint 2
  Use the same total ratio count for both parts, then test the two results independently.
Answer

$\begin{aligned}BD&=12.75\\DC&=17.25.\end{aligned}$

Full solution

The opposite-side parts have ratio $17:23$ , so the whole contains $40$ ratio parts.
$\begin{aligned}BD&=30\cdot\frac{17}{40},\\&=12.75,\\DC&=30\cdot\frac{23}{40},\\&=17.25.\end{aligned}$

The sum is $12.75+17.25=30$ , agreeing with the given whole.
The ratio is $12.75/17.25=17/23$ , agreeing with $AB/AC$ .

Answer

$\begin{aligned}BD&=12.75\\DC&=17.25.\end{aligned}$

Key idea

A whole-side denominator adds the two ratio counts.
Problem 3 Recover an adjacent side from the base parts

In triangle $ABC$ , $AD$ internally bisects $\angle BAC$ . The base parts are
$\begin{aligned}BD&=9\\DC&=15\end{aligned}$
, and the adjacent side $AC=20$ .

Find $AB$ . Explain why using $BD/BC$ in place of $BD/DC$ would change the equation.
1. Hint 1
  Write the part-to-part equality before inserting any numbers.
2. Hint 2
  The whole BC includes both base parts; it has a different denominator from DC.
Answer

$AB=12$ .

Full solution

The correct part-to-part equality gives
$\begin{aligned}\frac{AB}{20}&=\frac 9{15},\\AB&=12.\end{aligned}$

Here $BC=24$ , so $BD/BC=9/24$ is a different fraction.
It would match $AB/(AB+AC)$ , not $AB/AC$ .
The resulting outer sides 12, 20, 24 satisfy the triangle inequality.

Answer

$AB=12$ .

Key idea

Changing a part denominator to a whole changes the corresponding comparison.
Problem 4 Choose the actual bisector intersection

Triangle $ABC$ has
$\begin{aligned}AB&=15\\AC&=24\\BC&=26.\end{aligned}$
Two candidate interior points are specified by
$\begin{aligned}BD_1&=10\\BD_2&=11.\end{aligned}$
Which of $AD_1,AD_2$ is the internal angle bisector? Calculate the leftover parts and justify both decisions.
1. Hint 1
  Both candidates lie on the same base of fixed total length.
2. Hint 2
  The converse requires equality of ratios; an inequality in the necessary direct-theorem ratio rules a candidate out.
Answer

$AD_1$ is the internal bisector; $AD_2$ is not. The leftover parts are
$\begin{aligned}D_1C&=16\\D_2C&=15.\end{aligned}$

Full solution

Subtract from the given whole.
$\begin{aligned}D_1C&=26-10,\\&=16,\\D_2C&=26-11,\\&=15.\end{aligned}$

The adjacent-side ratio is $15/24=5/8$ .
Candidate $D_1$ gives $10/16=5/8$ , so the interior angle-bisector ratio converse proves $AD_1$ is the bisector.

Candidate $D_2$ gives $11/15$ , which differs from $5/8$ .
The direct theorem would require equality if it were a bisector, so $AD_2$ is not one.

Answer

$AD_1$ is the internal bisector; $AD_2$ is not. The leftover parts are
$\begin{aligned}D_1C&=16\\D_2C&=15.\end{aligned}$

Key idea

Use the converse to accept a ratio match and the direct theorem to rule out a mismatch.
Problem 5 Resolve an algebraic ratio constraint

In a triangle with internal bisector $AD$ , the lengths are
$\begin{aligned}AB&=2t+4\\AC&=3t+5\\BD&=9\\DC&=13.\end{aligned}$
Find $t$ , $AB$ , and $AC$ . Verify that the resulting outer side lengths are compatible with a nondegenerate triangle.

Use the named lengths, then check the outer triangle.

Text description of this figure

The given algebraic adjacent-side lengths and numerical base parts are labeled on a schematic internal-bisector figure.
1. Hint 1
  The base parts fix the adjacent-side ratio in the B-then-C order.
2. Hint 2
  After solving, test positivity and compare the largest outer side with the sum of the other two.
Answer

$\begin{aligned}t&=7\\AB&=18\\AC&=26\end{aligned}$
; together with $BC=22$ , these form a nondegenerate triangle.

Full solution

Use the exact ratio constraint.
$\begin{aligned}\frac{2t+4}{3t+5}&=\frac 9{13},\\13(2t+4)&=9(3t+5),\\26t+52&=27t+45,\\t&=7.\end{aligned}$

Substitution gives
$\begin{aligned}AB&=18\\AC&=26.\end{aligned}$
The whole base is
$\begin{aligned}BC&=9+13,\\&=22.\end{aligned}$
All are positive, and the largest side satisfies $26<18+22$ , so the triangle is nondegenerate.

The ratio checks:
$18/26=9/13.$

Answer

$\begin{aligned}t&=7\\AB&=18\\AC&=26\end{aligned}$
; together with $BC=22$ , these form a nondegenerate triangle.

Key idea

An algebraic answer must still describe positive compatible geometric lengths.
Problem 6 Explain the auxiliary equal length

Triangle $ABC$ has
$\begin{aligned}AB&=18\\AC&=25\\BC&=32.\end{aligned}$
Let $AD$ be its actual internal angle bisector. Through $C$ , draw the parallel to $AD$ meeting line $BA$ at $E$ .

(a) Explain why the order is $B,A,E$ and why $AE=25$ .

(b) Write the side-splitter proportion in triangle $BEC$ that gives the required base ratio. Explain why the denominator is $AE$ , not $BE$ .
1. Hint 1
  The points B,C are on opposite sides of line AD. The parallel through C cannot cross that line.
2. Hint 2
  Transfer the two half-angle equalities to E,C, then identify the first and leftover pieces on BE.
Answer

(a) $B,A,E$ occur in that order and
$\begin{aligned}AE&=AC,\\&=25.\end{aligned}$
(b)
$\begin{aligned}BA/AE&=BD/DC,\\&=18/25.\end{aligned}$

Full solution

Because $D$ is interior to $BC$ , the points $B,C$ lie on opposite sides of line $AD$ .
The parallel through $C$ stays on its side of $AD$ ; its intersection with line $BA$ therefore lies beyond $A$ , giving order $B,A,E$ .

Parallel-line angles give
$\begin{aligned}\angle AEC\cong\angle BAD\\\angle ACE\cong\angle CAD.\end{aligned}$
The actual bisector makes the two half-angles equal.
Thus the angles at $E,C$ in triangle $AEC$ are equal, so their opposite sides satisfy
$\begin{aligned}AE&=AC,\\&=25.\end{aligned}$

In triangle $BEC$ , the cut $AD$ is parallel to $EC$ , with both cut points interior.
Side-splitter compares first part with leftover part:
$\begin{aligned}BA/AE&=BD/DC,\\&=18/25.\end{aligned}$
Here $BE=43$ is the whole auxiliary side, so it cannot replace the leftover part $AE$ in this equation.

First prove the copied length; then use the parallel cut.

Answer

(a) $B,A,E$ occur in that order and
$\begin{aligned}AE&=AC,\\&=25.\end{aligned}$
(b)
$\begin{aligned}BA/AE&=BD/DC,\\&=18/25.\end{aligned}$

Key idea

The auxiliary isosceles triangle supplies exactly the leftover length needed by side-splitter.
Problem 7 Identify the true bisector without an angle measurement

Triangle $ABC$ has
$\begin{aligned}AB&=16\\AC&=27\\BC&=31.\end{aligned}$
A given interior point $D$ satisfies $BD:DC=16:27$ .

Construct the actual internal bisector conceptually, calling its intersection $D_0$ . Explain why $D$ and $D_0$ must coincide, rather than simply assuming the given ratio means equal angles.
1. Hint 1
  Apply the direct theorem to the actual bisector first.
2. Hint 2
  On one fixed segment, a positive ratio determines a unique first-part fraction of the whole.
Answer

$D=D_0$ , so $AD$ internally bisects $\angle BAC$ .

Full solution

The direct theorem applied to the actual internal bisector gives
$BD_0:D_0C=16:27.$
The given point has the same ratio, on the same side $BC$ of length $31$ .

Each first part must be $16/43$ of the whole.
$\begin{aligned}BD_0&=\frac{16}{43}\cdot 31,\\&=\frac{496}{43},\\BD&=BD_0.\end{aligned}$

Both points lie on segment $BC$ at the same distance from $B$ , so they are the same point.

It follows that the given segment $AD$ is the actual bisector.
This is the converse argument, not an unproved reversal of the theorem.

Uniqueness identifies the given point with the true bisector intersection.

Answer

$D=D_0$ , so $AD$ internally bisects $\angle BAC$ .

Key idea

The actual bisector and unique interior division bridge ratio data to equal angles.
Problem 8 Test the proposed small-triangle similarity

Triangle $ABC$ has
$\begin{aligned}AB&=10\\AC&=16\\BC&=22\end{aligned}$
, with internal angle bisector $AD$ .

Does $\triangle ABD\sim\triangle ACD$ hold with $A$ matching $A$ , $B$ matching $C$ , and $D$ matching $D$ ? Compare the side ratios in this specified order without estimating $AD$ . Explain why equal half-angles do not settle the question.

Follow the stated correspondence when comparing the shared side.

Text description of this figure

The schematic marks AD as the internal bisector and labels the unequal adjacent sides 10 and 16. AD is shared by the two small triangles.
1. Hint 1
  The common segment AD matches itself in this proposed order.
2. Hint 2
  A similarity would require every target-to-original side ratio to agree.
Answer

No. The proposed correspondence gives $AC/AB=1.6$ but $AD/AD=1$ .

Full solution

For the stated correspondence, side $AB$ matches $AC$ , while $AD$ matches itself.
Their target-to-original ratios differ.
$\begin{aligned}\frac{AC}{AB}&=\frac{16}{10},\\&=1.6,\\\frac{AD}{AD}&=1.\end{aligned}$

Thus this ordered similarity is impossible.
The equal half-angles provide only one equal-angle pair and do not justify AA.

The ratio $AB/AC$ in the bisector theorem compares adjacent sides of the original triangle; it is not the scale factor between these two smaller triangles.

Answer

No. The proposed correspondence gives $AC/AB=1.6$ but $AD/AD=1$ .

Key idea

A within-triangle ratio need not be a similarity scale factor.
Problem 9 Combine a ratio with a side difference

In triangle $ABC$ , $AD$ is an internal bisector with
$\begin{aligned}BD&=6\\DC&=8.\end{aligned}$
The adjacent side $AC$ is $5$ units longer than $AB$ .

Find $AB$ and $AC$ , then check whether they can form a nondegenerate triangle with the given base.
1. Hint 1
  The two base parts determine the ratio of the adjacent sides.
2. Hint 2
  Express both adjacent sides using one common positive ratio unit; use their difference to find that unit.
Answer

$\begin{aligned}AB&=15\\AC&=20\end{aligned}$
; the base is $14$ and the triangle is nondegenerate.

Full solution

The adjacent-side ratio is
$\begin{aligned}AB:AC&=6:8,\\&=3:4.\end{aligned}$
Write the lengths as $3u$ and $4u$ .
Their difference is one ratio unit, so $u=5$ .

Therefore
$\begin{aligned}AB&=15\\AC&=20.\end{aligned}$
The base is
$\begin{aligned}BC&=6+8,\\&=14.\end{aligned}$
All lengths are positive, and the largest side satisfies $20<15+14$ , so a nondegenerate triangle is possible.

The side ratio is $15/20=3/4$ , matching the part ratio $6/8$ .

Check the ratio and the triangle after using the difference constraint.

Answer

$\begin{aligned}AB&=15\\AC&=20\end{aligned}$
; the base is $14$ and the triangle is nondegenerate.

Key idea

A side difference can determine the common unit hidden in a ratio.
Problem 10 Assess a triangle record

A record lists
$\begin{aligned}AB&=7\\AC&=12\\BC&=4\end{aligned}$
, then proposes an internal bisector $AD$ with $BD:DC=7:12$ .

(a) Compute the positive base parts that this proportion would give arithmetically.

(b) Decide whether all the recorded measurements can describe a nondegenerate triangle. Explain which earlier condition decides the issue.
1. Hint 1
  Divide the base arithmetically into nineteen ratio parts.
2. Hint 2
  A ratio calculation assumes a triangle exists. Test the outer side lengths separately.
Answer

(a)
$\begin{aligned}BD&=28/19\\DC&=48/19.\end{aligned}$
(b) No such triangle exists, since $7+4<12$ .

Full solution

The arithmetic division is positive and adds to $4$ .
$\begin{aligned}BD&=4\cdot\frac 7{19},\\&=\frac{28}{19},\\DC&=4\cdot\frac{12}{19},\\&=\frac{48}{19}.\end{aligned}$

However, the outer side lengths fail the triangle inequality:
$\begin{aligned}AB+BC&=7+4,\\&=11<12,\\&=AC.\end{aligned}$
They cannot describe a triangle.

The ratio theorem applies to an existing nondegenerate triangle.
Positive formal parts alone do not establish that geometric premise.

Answer

(a)
$\begin{aligned}BD&=28/19\\DC&=48/19.\end{aligned}$
(b) No such triangle exists, since $7+4<12$ .

Key idea

Check existence conditions independently of a successful proportional calculation.

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