Triangle Angle Bisectors worksheet with problems and diagrams.

Let’s solve each problem step by step.

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Problem 1:
We’re told that m∠DFR = 56°, and we need to find m∠1.
Looking at the triangle, angle DFR is at vertex F. The line drawn from F splits angle DFR into two equal parts — because it’s an angle bisector.
So, ∠1 is half of ∠DFR.
→ m∠1 = 56° ÷ 2 = 28°

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Problem 2:
We’re told m∠1 = 25°, and we need to find m∠NXE.
In triangle NXE, the line from X splits angle NXE into two equal angles: ∠1 and ∠2.
Since it’s an angle bisector, ∠1 = ∠2 = 25°.
So, the full angle NXE = ∠1 + 2 = 25° + 25° = 50°

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Problem 3:
Each triangle shows its three angle bisectors intersecting at point C (the incenter).
We’re told CI = 8 and CZ = 17. We need to find ZI.
Points Z, C, I are on a straight line (since they lie on the same angle bisector from vertex Z).
So, ZI = ZC + CI = 17 + 8 = 25

Wait — let’s check the diagram labeling. In triangle ZRT, point C is inside, and ZI is part of the bisector from Z to side RT. Point I is on side RT? Actually, looking at standard notation:
In problem 3, points U, I, H are feet of perpendiculars? No — actually, since these are angle bisectors meeting at C, then segments like CI and CZ are parts of the bisectors.
But if Z, C, I are colinear with C between Z and I, then ZI = ZC + CI.
Given CZ = 17, CI = 8 → ZI = 17 + 8 = 25

Yes, that makes sense.

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Problem 4:
NP = 13, CN = 6. Find CP.
Again, C is the incenter. Points N, C, P are on the same angle bisector (from vertex L down to side EP).
Assuming C is between N and P, then NP = NC + CP.
So, 13 = 6 + CP → CP = 13 - 6 = 7

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Problem 5:
JY = 10, CJ = 4. Find CY.
Points J, C, Y are on the same angle bisector (from vertex U to side PY).
If C is between J and Y, then JY = JC + CY.
So, 10 = 4 + CY → CY = 10 - 4 = 6

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Problem 6:
QJ = 16, CQ = 4. Find CJ.
Points Q, C, J are on the same angle bisector (from vertex P to side TJ).
If C is between Q and J, then QJ = QC + CJ.
So, 16 = 4 + CJ → CJ = 16 - 4 = 12

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Problem 7:
CJ = 18. Find CX.
This one is tricky without more info — but wait! In all these diagrams, point C is the incenter, and the lines from vertices to opposite sides are angle bisectors. But also, in some cases, there are perpendiculars drawn (like in problems 3–8, you see right-angle marks at the base). That suggests those are not just angle bisectors, but also altitudes or something else? Wait — no, the title says “Triangle Angle Bisectors”, so even though there are right-angle symbols, they might be indicating where the bisector meets the side perpendicularly? Not necessarily.

Actually, re-examining: in problems 3–8, the triangles show three angle bisectors intersecting at C, and also have perpendiculars dropped from C to the sides (with right-angle marks), meaning C is equidistant to all sides — which is true for the incenter. But the lengths given are along the bisectors themselves.

In problem 7: CJ = 18. Find CX.
Looking at the diagram: triangle PYM, with angle bisectors from P, Y, M meeting at C. From Y, the bisector goes to side PM, hitting at X. So Y-C-X is a straight line.
Similarly, from P, the bisector goes to YM, hitting at J. So P-C-J is a straight line.
But we’re only given CJ = 18. How does that relate to CX?

Wait — perhaps in this context, since C is the incenter, and the segments from C to the sides are perpendicular (as shown by right angles), but the lengths given (like CJ, CX) are distances from C to the sides? But the problem says “CJ = 18” — and in the diagram, J is on side YM, and CJ is perpendicular to YM (right angle mark). Similarly, CX is perpendicular to PM.

Ah! Here’s the key: The distance from the incenter to any side is the same — that’s the inradius. So if CJ is the perpendicular distance from C to side YM, and CX is the perpendicular distance from C to side PM, then CJ = CX.

Therefore, if CJ = 18, then CX = 18

That must be it. The right-angle marks indicate perpendiculars, so CJ and CX are both radii of the incircle → equal length.

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Problem 8:
CM = 3, CA = 11. Find AM.
Points A, C, M are on the same angle bisector (from vertex A to side RX).
Assuming C is between A and M, then AM = AC + CM = 11 + 3 = 14

Wait — let’s confirm the order. In triangle ARX, angle bisector from A hits side RX at M. Point C is the incenter, lying on AM. So yes, A-C-M are colinear, with C between A and M.
So AM = AC + CM = 11 + 3 = 14

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Now, let’s double-check each answer:

1) 56° / 2 = 28° ✔
2) 25° × 2 = 50° ✔
3) 17 + 8 = 25 ✔
4) 13 - 6 = 7 ✔
5) 10 - 4 = 6 ✔
6) 16 - 4 = 12 ✔
7) Since CJ and CX are both perpendicular distances from incenter to sides → equal → 18 ✔
8) 11 + 3 = 14 ✔

All seem correct.

Final Answer:
1) 28
2) 50
3) 25
4) 7
5) 6
6) 12
7) 18
8) 14