Geometry path · G06
Before this lesson: Triangle area ratios, The midpoint theorem
Your goal: Use parallelism or an angle bisector to obtain a correctly oriented ratio.
Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.
Names you may know: basic proportionality theorem; intercept theorem; angle bisector theorem.
Parallel lines and angle bisectors: the key idea
If D lies on AB and E on AC with DE∥BC, then AD/AB=AE/AC=DE/BC and AD/DB=AE/EC. Conversely, an equal division ratio on the two sides gives a parallel connecting line. The internal angle-bisector theorem says AD/DC=AB/BC when the bisector from B meets AC at D. Its external version uses the same ratio in absolute segment lengths on the side line, with an external division point; when the adjacent sides are equal the external bisector is parallel to the opposite side and has no finite meeting point.
A worked example
In triangle ABC, D∈AB,E∈AC, DE∥BC. If AD=3,DB=2 and AC=10, find AE.
AB=5, so the similarity scale AD/AB is 3/5. Hence AE=(3/5)·10=6. Using AD/DB=3/2 as the whole-side scale would be a mistake.
Your turn: change one thing
An internal bisector from A meets BC at D. If AB=6,AC=9 and BC=10, find BD.
Try this on paper before opening the explanation.
Compare your reasoning
BD/DC=AB/AC=2/3. Therefore BD=(2/5)·10=4 and DC=6.
Pause and check
A trap to avoid: Reversing one ratio or ignoring a point on an extension.
Practise and adjust the level
Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.
Interactive practice loads here. You can also use the complete question set below.
Write a complete argument
Prove the internal angle-bisector theorem using areas.
Planning hint
List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.
Read the full solution after your attempt
Let AD bisect ∠A in triangle ABC. Triangles ABD and ACD have area ratio BD/DC by their common altitude to BC. They also have ratio [AB·AD·sin∠BAD]/[AC·AD·sin∠DAC]=AB/AC because the two angles are equal. Equating the ratios yields BD/DC=AB/AC.
My proof notebook
Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.
All nine practice questions
Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.
1. If DE∥BC, AD/AB equals which ratio?
Foundation
- AE/EC
- DB/AB
- BC/DE
- AE/AC
Hint
Use corresponding sides of similar triangles.
Answer and reasoning
AE/AC. The scale is measured from A to each whole side.
2. An internal angle bisector divides the opposite side in what ratio?
Foundation
- The adjacent side lengths
- Always 1:1
- The squared side lengths
- The opposite angles only
Hint
Use the angle-bisector theorem.
Answer and reasoning
The adjacent side lengths. BD/DC=AB/AC for the bisector from A.
3. AD=3,DB=2,AC=10 and DE∥BC. What is AE?
Core
- 15
- 4
- 5
- 6
Hint
AD/AB=3/5.
Answer and reasoning
6. AE=10·3/5=6.
4. AB=6,AC=9,BC=10 and AD bisects ∠A. What is BD?
Core
- 4
- 5
- 6
- 3
Hint
Split 10 in ratio 2:3.
Answer and reasoning
4. BD=10·2/5=4.
5. What error occurs when using AD/DB as AD/AB?
Stretch
- No error
- Only units change
- The ratio becomes negative always
- A part-to-part ratio is confused with part-to-whole
Hint
The denominators describe different segments.
Answer and reasoning
A part-to-part ratio is confused with part-to-whole. AB=AD+DB, so the two ratios generally differ.
6. If adjacent sides are equal, the external angle bisector is what relative to the opposite side?
Stretch
- The same line
- A median inside the triangle
- Parallel
- Perpendicular always
Hint
The internal bisector is the symmetry axis.
Answer and reasoning
Parallel. The external bisector is perpendicular to the internal one and parallel to the base in this isosceles case.
7. The internal angle bisector splits which angle into equal parts?
Foundation
- The angle at its starting vertex
- Every triangle angle
- The opposite side angle
- A straight angle always
Hint
Identify the ray from the vertex.
Answer and reasoning
The angle at its starting vertex. Its two adjacent angles inside the triangle are equal.
8. AB=4,AC=6 and BC=5. If AD bisects ∠A, what is BD?
Core
- 4
- 1
- 2
- 3
Hint
Use BD:DC=2:3.
Answer and reasoning
2. BD is 2/5 of 5, hence 2.
9. In a proportionality problem, why keep side labels attached to ratios?
Stretch
- Labels make lengths integers
- Reversing one ratio alone can change the equation
- Ratios have no direction
- All ratios equal one
Hint
Correspondence determines numerator and denominator order.
Answer and reasoning
Reversing one ratio alone can change the equation. Matching divisions must be oriented consistently.
Choose your next step
Continue to Similar triangles. If this felt difficult, return to a prerequisite above. Every lesson stays open.
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