Geometry Worksheets | Triangle Worksheets - Free Printable

Let’s solve each problem one by one, step by step. We’ll use the properties of angle bisectors and triangle centers (like incenter) where needed.

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Problem 1:
Given: m∠DFR = 55°, and we’re told that an angle bisector is drawn. Looking at the diagram, ∠1 and ∠2 are parts of ∠DFR, and since it’s an angle bisector, it splits ∠DFR into two equal angles. So ∠1 = ∠2.

Therefore:
m∠1 = 55° ÷ 2 = 27.5°

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Problem 2:
Given: m∠1 = 25°, and we need to find m∠NXE.
Looking at the diagram, ∠1 and ∠2 are the two halves of ∠NXE, split by the angle bisector from X. So ∠NXE = ∠1 + ∠2 = 25° + 25° = 50°

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Problem 3:
Each triangle shows its three angle bisectors intersecting at point C — this means C is the incenter. In a triangle, the incenter is equidistant from all sides, but here we’re given lengths along the bisectors.

Wait — actually, looking at the labels: CI = 8, CZ = 17, and we need ZI.

Points Z, C, I lie on the same line (the angle bisector from Z). Since C is between Z and I? Or is I between Z and C?

Looking at the diagram: Point C is inside the triangle, and Z is a vertex, I is on the opposite side. So the segment goes Z → C → I.

So ZI = ZC + CI = 17 + 8 = 25

But wait — the problem says “CI = 8 and CZ = 17”. If C is between Z and I, then yes, ZI = ZC + CI = 17 + 8 = 25.

✔ Confirmed.

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Problem 4:
NP = 13, CN = 6, find CP.

Again, C is the incenter. Points N, C, P are colinear? Looking at the diagram: N is on base EP, P is a vertex, C is inside. The line is from P through C to N.

So if NP = 13, and CN = 6, then CP = NP - CN = 13 - 6 = 7

Because C lies between P and N.

✔ Correct.

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Problem 5:
JY = 10, CJ = 4, find CY.

Points J, C, Y are on the same line (angle bisector from Y). C is between J and Y? Or J between C and Y?

Diagram: Y is vertex, J is on opposite side, C is inside. So order is Y → C → J.

Thus, YJ = YC + CJ → 10 = CY + 4 → CY = 10 - 4 = 6

✔ Yes.

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Problem 6:
QJ = 16, CQ = 4, find CJ.

Points Q, C, J are on the same line. Q is on side TJ, J is vertex, C is inside. So order is J → C → Q.

Thus, JQ = JC + CQ → 16 = CJ + 4 → CJ = 16 - 4 = 12

✔ Correct.

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Problem 7:
CJ = 18, find CX.

Looking at diagram: C is incenter. J and X are points on different sides. But notice: both CJ and CX are segments from incenter C to the sides — and in a triangle, the incenter is equidistant to all sides! That distance is the radius of the incircle.

But wait — are CJ and CX perpendicular to the sides? In the diagram, they appear to be marked with right angles — meaning they are the perpendicular distances from C to the sides.

In that case, since C is the incenter, the perpendicular distances to all sides are equal.

So if CJ = 18, and CX is also a perpendicular from C to another side, then CX = CJ = 18

✔ Yes — because incenter is equidistant to all sides.

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Problem 8:
CM = 3, CA = 11, find AM.

Points A, C, M are on the same line — angle bisector from A. C is incenter, so between A and M.

Order: A → C → M

So AM = AC + CM = 11 + 3 = 14

✔ Correct.

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Final Answer:

1) 27.5
2) 50
3) 25
4) 7
5) 6
6) 12
7) 18
8) 14