Geometry Worksheets | Triangle Worksheets - Free Printable

This worksheet is about angle bisectors in triangles, with two main concepts:

1. A single angle bisector splits an angle into two equal parts.
2. The three angle bisectors of a triangle intersect at a single point called the incenter (C), and this point is equidistant from all three sides of the triangle. This distance is the radius of the incircle.

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Let’s solve each problem step by step.

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Problem 1:

> m∠DFR = 55°. Find m∠1.

- The line drawn from F to side DR is the angle bisector of ∠DFR.
- So it splits ∠DFR into two equal angles: ∠1 and ∠2.
- Therefore, m∠1 = 55° ÷ 2 = 27.5°

✔ Answer: 27.5°

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Problem 2:

> Find m∠NXE if m∠1 = 25°

- The line XC is the angle bisector of ∠NXE.
- So ∠1 and ∠2 are equal, and together they make up ∠NXE.
- So m∠NXE = ∠1 + ∠2 = 25° + 25° = 50°

✔ Answer: 50°

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Problems 3–8:

These involve the incenter (C) — the point where all three angle bisectors meet. A key property:

> The incenter is equidistant from all three sides of the triangle.

That means the perpendicular distances from C to each side are equal. These distances are shown as segments like CI, CZ, CP, etc., which are perpendicular to the sides (indicated by right-angle symbols).

So, for example, in Problem 3:
> CI = 8 and CZ = 17. Find ZI.

Wait — this doesn’t make sense if CI and CZ are both distances from C to the sides — they should be equal!

But looking at the diagram:

- In triangle ZRT, point C is the incenter.
- CI is perpendicular to side ZT → so CI is the distance from C to side ZT.
- CZ is perpendicular to side RT → so CZ is the distance from C to side RT.

But if C is the incenter, then CI = CZ = CH (all perpendicular distances to the sides must be equal). So CI = CZ = 8? But it says CI=8 and CZ=17 — contradiction?

Actually — let’s look again.

In Problem 3:
> CI = 8 and CZ = 17. Find ZI.

Looking at the diagram, ZI is not a perpendicular segment — it's the segment from vertex Z to point I on side ZT.

Wait — perhaps we misread. Let’s check the diagram carefully.

Actually, in these problems, C is the incenter, and the segments like CI, CZ, CP, etc., are perpendiculars from C to the sides, so they represent the inradius — and thus they must all be equal.

So if CI = 8 and CZ = 17, that contradicts the definition — unless...

Ah! Wait — maybe CI and CZ are NOT both perpendicular distances? Let’s check the diagram.

Looking at Problem 3’s diagram:

- Point I is on side ZT, and CI is perpendicular to ZT → so CI is the inradius.
- Point Z is a vertex — CZ is from vertex Z to point C — that’s not a perpendicular to a side!

Oh! That’s the mistake.

In Problems 3–8, the notation is:

- CI, CZ, CP, etc. — some are distances from C to sides (perpendiculars), others are segments from vertices to C.

But in the problem statements, they say:

> “Each triangle shows its three angle bisectors intersecting at point C.”

And in the diagrams, we see right-angle marks at points like I, H, U, N, Q, X, B — meaning those are the feet of perpendiculars from C to the sides.

So:

- CI, CH, CU — in Problem 3 — are perpendiculars → so they are equal → inradius.
- But CZ — is from vertex Z to C — that’s not a perpendicular.

Wait — in Problem 3: “CI = 8 and CZ = 17. Find ZI.”

Looking at the diagram: ZI is part of side ZT, from vertex Z to point I (where perpendicular from C meets ZT).

So triangle CZI is a right triangle at I.

Because CI ⊥ ZT → ∠CIZ = 90°.

So in right triangle CZI:

- CI = 8 (one leg)
- CZ = 17 (hypotenuse)
- ZI = ? (other leg)

Use Pythagoras:

> ZI² + CI² = CZ²
> ZI² + 8² = 17²
> ZI² + 64 = 289
> ZI² = 289 - 64 = 225
> ZI = √225 = 15

✔ Answer: 15

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Problem 4:

> NP = 13 and CN = 6. Find CP.

Diagram: Triangle LEP, C is incenter.

CN is perpendicular to EP → so CN = 6 is the inradius.

NP = 13 — that’s from N (foot of perpendicular on EP) to P (vertex).

CP is from C to P — so triangle CNP is right-angled at N.

So in right triangle CNP:

- CN = 6 (leg)
- NP = 13 (leg)
- CP = hypotenuse

Pythagoras:

> CP² = CN² + NP² = 6² + 13² = 36 + 169 = 205
> CP = √205

We can leave it as √205 or approximate, but since others are integers, probably exact form.

✔ Answer: √205

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Problem 5:

> JY = 10 and CJ = 4. Find CY.

Diagram: Triangle PUY, C is incenter.

CJ is perpendicular to PY → so CJ = 4 is inradius.

JY = 10 — from J (foot on PY) to Y (vertex).

CY is from C to Y — so triangle CJY is right-angled at J.

So:

> CY² = CJ² + JY² = 4² + 10² = 16 + 100 = 116
> CY = √116 = √(4×29) = 2√29

✔ Answer: 2√29

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Problem 6:

> QJ = 16 and CQ = 4. Find CJ.

Diagram: Triangle PKJ, C is incenter.

CQ is perpendicular to PJ → CQ = 4 (inradius).

QJ = 16 — from Q (foot on PJ) to J (vertex).

CJ is from C to J — so triangle CQJ is right-angled at Q.

So:

> CJ² = CQ² + QJ² = 4² + 16² = 16 + 256 = 272
> CJ = √272 = √(16×17) = 4√17

✔ Answer: 4√17

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Problem 7:

> CJ = 18. Find CX.

Diagram: Triangle PYM, C is incenter.

CJ is perpendicular to YM → CJ = 18 (inradius).

CX is perpendicular to PM → so CX should also be the inradius → CX = CJ = 18

Because incenter is equidistant to all sides.

✔ Answer: 18

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Problem 8:

> CM = 3 and CA = 11. Find AM.

Diagram: Triangle RAX, C is incenter.

CM is perpendicular to RX → CM = 3 (inradius).

CA is from C to A — A is on side RX? Looking at diagram: A is foot of perpendicular from C to RX? No — wait.

Actually, in diagram:

- CM is perpendicular to RX → M is foot → CM = 3
- CA is from C to A — and A is on side RX? Yes, and there’s a right angle at A? No — in diagram, CA is not perpendicular — wait.

Wait — looking at diagram: There’s a right angle at M and at B and at Y.

In triangle RAX:

- CM ⊥ RX → M is foot → CM = 3
- CA is from C to A — but A is on side RX? And is CA perpendicular? No — the right angle is at M, not A.

Actually, in diagram, point A is on side RX, and CA is drawn, but no right angle at A.

Wait — perhaps A is the foot? But the right angle mark is at M.

Re-examining: In Problem 8, it says “CM = 3 and CA = 11. Find AM.”

In diagram, C is incenter, M is foot on RX, A is another point on RX.

So points R-A-M-X are on side RX? Probably R-A-M-X in order.

Then AM is the distance between A and M on side RX.

But we don’t know positions.

Wait — perhaps triangle CAM is right-angled?

No — only CM is perpendicular to RX.

So CA is a line from C to A on RX, and CM is perpendicular from C to RX.

So triangle CMA is right-angled at M.

Because CM ⊥ RX, and A is on RX, so CM ⊥ AM.

Thus, triangle CMA is right-angled at M.

So:

- CM = 3 (leg)
- CA = 11 (hypotenuse)
- AM = ? (other leg)

Pythagoras:

> 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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## 🔑 Key Concepts Used:

- Angle bisector divides an angle into two equal parts.
- Incenter is the intersection of angle bisectors and is equidistant from all sides (inradius).
- When you have a right triangle formed by the incenter, a vertex, and the foot of the perpendicular, use the Pythagorean theorem.

Let me know if you want diagrams explained further!