Triangle Angle Bisectors worksheet with problems and diagrams for calculating angles and segment lengths.
Worksheet titled "Triangle Angle Bisectors" with eight problems involving finding angles and lengths using angle bisectors in triangles, featuring diagrams and mathematical expressions.
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Show Answer Key & Explanations
Step-by-step solution for: Geometry Worksheets | Triangle Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Geometry Worksheets | Triangle Worksheets
Let’s solve each problem step by step, using the properties of angle bisectors in triangles.
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## 🔷 Key Concepts:
- An angle bisector divides an angle into two equal parts.
- So if a ray bisects ∠A, then each half-angle is (∠A)/2.
- When all three angle bisectors of a triangle intersect at point C, that point is called the incenter.
- The incenter is equidistant from all three sides of the triangle.
- This distance is the inradius, and it's measured perpendicularly to each side.
- Therefore, the perpendicular distances from the incenter to each side are equal.
> 💡 Important: In problems 3–8, since C is the incenter, the perpendicular segments from C to the sides (like CI, CZ, CP, etc.) are all equal in length — they are the inradius.
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## ✔ Problem-by-Problem Solutions:
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- The bisector splits ∠DFR into two equal angles.
- So, ∠1 = ∠2 = 55° ÷ 2 = 27.5°
✔ Answer: 27.5°
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- Here, XC bisects ∠NXE, so ∠1 and ∠2 are equal.
- So, ∠NXE = ∠1 + ∠2 = 25° + 25° = 50°
✔ Answer: 50°
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⚠️ Wait — this seems impossible under normal geometry rules.
- If C is the incenter, then CI and CZ must be equal, because both are perpendicular distances from C to the sides (inradius).
- But here, CI = 8 and CZ = 17 → contradiction.
➡️ Likely typo or mislabeling.
But looking at the diagram:
- Point I is on side ZT
- Point Z is a vertex
- CZ is drawn from vertex Z to incenter C — not the perpendicular distance!
Ah! Important distinction:
- In diagrams 3–8, the segments labeled like CI, CZ, CP, etc., are NOT always the perpendicular distances.
- Some are from vertex to incenter, others are perpendicular to sides.
Looking closely at the diagrams:
In problem 3:
- CI is drawn perpendicular to side ZT (marked with right angle)
- CZ is drawn from vertex Z to incenter C — this is NOT the inradius
So we have:
- CI = 8 → this is the inradius (distance from C to side ZT)
- CZ = 17 → this is the length from vertex Z to incenter C
We are to find ZI — which is part of side ZT.
Since CI ⊥ ZT, triangle CZI is a right triangle with:
- Right angle at I
- CI = 8 (leg)
- CZ = 17 (hypotenuse)
Use Pythagoras:
> ZI² + CI² = CZ²
> ZI² + 8² = 17²
> ZI² + 64 = 289
> ZI² = 225
> ZI = √225 = 15
✔ Answer: 15
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- CN is perpendicular to EP (right angle marked) → CN = 6 is the inradius
- NP = 13 is the length along side EP from N to P
- CP is from incenter C to vertex P
We need to find CP.
Again, triangle CNP is right-angled at N:
- CN = 6 (leg)
- NP = 13 (other leg)
- CP = hypotenuse
Pythagoras:
> CP² = CN² + NP² = 6² + 13² = 36 + 169 = 205
> CP = √205 ≈ 14.32 — but since the problem likely expects exact value,
✔ Answer: √205
*(Note: If decimal expected, ≈14.3, but exact form is better unless specified)*
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- CJ is perpendicular to PY (right angle at J) → CJ = 4 is inradius
- JY = 10 is part of side PY
- CY is from incenter C to vertex Y
Triangle CJY is right-angled at J:
- CJ = 4 (leg)
- JY = 10 (leg)
- CY = hypotenuse
> CY² = 4² + 10² = 16 + 100 = 116
> CY = √116 = √(4×29) = 2√29
✔ Answer: 2√29
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- CQ is perpendicular to TJ (right angle at Q) → CQ = 4 is inradius
- QJ = 16 is part of side TJ
- CJ is from incenter C to vertex J
Triangle CQJ is right-angled at Q:
- CQ = 4 (leg)
- QJ = 16 (leg)
- CJ = hypotenuse
> CJ² = 4² + 16² = 16 + 256 = 272
> CJ = √272 = √(16×17) = 4√17
✔ Answer: 4√17
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- CJ is perpendicular to side PM? Wait — let’s check diagram.
In problem 7:
- CJ is perpendicular to side YM (at J) → CJ = 18 is inradius
- CX is perpendicular to side PM (at X) → since C is incenter, CX = CJ = 18
Because incenter is equidistant to all sides → CX = CJ = 18
✔ Answer: 18
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- CM is perpendicular to side RX (at M) → CM = 3 is inradius
- CA is from vertex A to incenter C → CA = 11
- We need to find AM — which is part of side RX
Triangle CMA is right-angled at M:
- CM = 3 (leg)
- CA = 11 (hypotenuse)
- AM = other leg
> AM² + CM² = CA²
> AM² + 3² = 11²
> AM² + 9 = 121
> AM² = 112
> AM = √112 = √(16×7) = 4√7
✔ Answer: 4√7
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## 📝 Final Answers:
1) 27.5°
2) 50°
3) 15
4) √205
5) 2√29
6) 4√17
7) 18
8) 4√7
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Let me know if you’d like these boxed or formatted for printing!
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## 🔷 Key Concepts:
1. Angle Bisector (Single)
- An angle bisector divides an angle into two equal parts.
- So if a ray bisects ∠A, then each half-angle is (∠A)/2.
2. Incenter (Three Angle Bisectors Intersecting)
- When all three angle bisectors of a triangle intersect at point C, that point is called the incenter.
- The incenter is equidistant from all three sides of the triangle.
- This distance is the inradius, and it's measured perpendicularly to each side.
- Therefore, the perpendicular distances from the incenter to each side are equal.
> 💡 Important: In problems 3–8, since C is the incenter, the perpendicular segments from C to the sides (like CI, CZ, CP, etc.) are all equal in length — they are the inradius.
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## ✔ Problem-by-Problem Solutions:
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1) m∠DFR = 55°. Find m∠1.
- The bisector splits ∠DFR into two equal angles.
- So, ∠1 = ∠2 = 55° ÷ 2 = 27.5°
✔ Answer: 27.5°
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2) Find m∠NXE if m∠1 = 25°
- Here, XC bisects ∠NXE, so ∠1 and ∠2 are equal.
- So, ∠NXE = ∠1 + ∠2 = 25° + 25° = 50°
✔ Answer: 50°
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3) CI = 8 and CZ = 17. Find ZI.
⚠️ Wait — this seems impossible under normal geometry rules.
- If C is the incenter, then CI and CZ must be equal, because both are perpendicular distances from C to the sides (inradius).
- But here, CI = 8 and CZ = 17 → contradiction.
➡️ Likely typo or mislabeling.
But looking at the diagram:
- Point I is on side ZT
- Point Z is a vertex
- CZ is drawn from vertex Z to incenter C — not the perpendicular distance!
Ah! Important distinction:
- In diagrams 3–8, the segments labeled like CI, CZ, CP, etc., are NOT always the perpendicular distances.
- Some are from vertex to incenter, others are perpendicular to sides.
Looking closely at the diagrams:
In problem 3:
- CI is drawn perpendicular to side ZT (marked with right angle)
- CZ is drawn from vertex Z to incenter C — this is NOT the inradius
So we have:
- CI = 8 → this is the inradius (distance from C to side ZT)
- CZ = 17 → this is the length from vertex Z to incenter C
We are to find ZI — which is part of side ZT.
Since CI ⊥ ZT, triangle CZI is a right triangle with:
- Right angle at I
- CI = 8 (leg)
- CZ = 17 (hypotenuse)
Use Pythagoras:
> ZI² + CI² = CZ²
> ZI² + 8² = 17²
> ZI² + 64 = 289
> ZI² = 225
> ZI = √225 = 15
✔ Answer: 15
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4) NP = 13 and CN = 6. Find CP.
- CN is perpendicular to EP (right angle marked) → CN = 6 is the inradius
- NP = 13 is the length along side EP from N to P
- CP is from incenter C to vertex P
We need to find CP.
Again, triangle CNP is right-angled at N:
- CN = 6 (leg)
- NP = 13 (other leg)
- CP = hypotenuse
Pythagoras:
> CP² = CN² + NP² = 6² + 13² = 36 + 169 = 205
> CP = √205 ≈ 14.32 — but since the problem likely expects exact value,
✔ Answer: √205
*(Note: If decimal expected, ≈14.3, but exact form is better unless specified)*
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5) JY = 10 and CJ = 4. Find CY.
- CJ is perpendicular to PY (right angle at J) → CJ = 4 is inradius
- JY = 10 is part of side PY
- CY is from incenter C to vertex Y
Triangle CJY is right-angled at J:
- CJ = 4 (leg)
- JY = 10 (leg)
- CY = hypotenuse
> CY² = 4² + 10² = 16 + 100 = 116
> CY = √116 = √(4×29) = 2√29
✔ Answer: 2√29
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6) QJ = 16 and CQ = 4. Find CJ.
- CQ is perpendicular to TJ (right angle at Q) → CQ = 4 is inradius
- QJ = 16 is part of side TJ
- CJ is from incenter C to vertex J
Triangle CQJ is right-angled at Q:
- CQ = 4 (leg)
- QJ = 16 (leg)
- CJ = hypotenuse
> CJ² = 4² + 16² = 16 + 256 = 272
> CJ = √272 = √(16×17) = 4√17
✔ Answer: 4√17
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7) CJ = 18. Find CX.
- CJ is perpendicular to side PM? Wait — let’s check diagram.
In problem 7:
- CJ is perpendicular to side YM (at J) → CJ = 18 is inradius
- CX is perpendicular to side PM (at X) → since C is incenter, CX = CJ = 18
Because incenter is equidistant to all sides → CX = CJ = 18
✔ Answer: 18
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8) CM = 3 and CA = 11. Find AM.
- CM is perpendicular to side RX (at M) → CM = 3 is inradius
- CA is from vertex A to incenter C → CA = 11
- We need to find AM — which is part of side RX
Triangle CMA is right-angled at M:
- CM = 3 (leg)
- CA = 11 (hypotenuse)
- AM = other leg
> AM² + CM² = CA²
> AM² + 3² = 11²
> AM² + 9 = 121
> AM² = 112
> AM = √112 = √(16×7) = 4√7
✔ Answer: 4√7
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## 📝 Final Answers:
1) 27.5°
2) 50°
3) 15
4) √205
5) 2√29
6) 4√17
7) 18
8) 4√7
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Let me know if you’d like these boxed or formatted for printing!
Parent Tip: Review the logic above to help your child master the concept of worksheet altitude median angle bisector perpendicular bisector.
