Comprehensive geometry worksheet featuring 15 exercises on finding unknown values using the angle bisector theorem.

We are told that →YZ is the angle bisector in each diagram. That means it splits the angle at point Y into two equal angles.

We’ll solve each problem one by one, using the definition of an angle bisector:
> The two angles created by the bisector are equal.

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Problem 1

Given: ∠AYZ = 18°, and →YZ bisects ∠AYL
So, ∠LYZ = ∠AYZ = 18°
Therefore, total angle ∠AYL = 18° + 18° = 36°

✔ Answer: 36°

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Problem 2

Given: ∠ZYG = 39°, and →YZ bisects ∠SYG
So, ∠SYZ = ∠ZYG = 39°
Total angle ∠SYG = 39° + 39° = 78°

✔ Answer: 78°

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Problem 3

Given: ∠TYZ = 54°, and →YZ bisects ∠TYH
So, ∠HZY = ∠TYZ = 54°
Total angle ∠TYH = 54° + 54° = 108°

✔ Answer: 108°

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Problem 4

Given: ∠DYU = 140° (the large angle), and →YZ bisects ∠DYU
So, each half = 140° ÷ 2 = 70°

✔ Answer: 70°

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Problem 5

Given: ∠EYZ = 88°, and →YZ bisects ∠EYQ
So, ∠ZYQ = ∠EYZ = 88°
Total angle ∠EYQ = 88° + 88° = 176°

✔ Answer: 176°

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Problem 6

Given: ∠AYZ = 5x, ∠LYZ = 15°, and →YZ bisects ∠AYL
So, 5x = 15
→ x = 15 ÷ 5 = 3

✔ Answer: x = 3

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Problem 7

Given: ∠ZYG = 21°, ∠SYZ = 5x - 9°, and →YZ bisects ∠SYG
So, 5x - 9 = 21
→ 5x = 30
→ x = 6

✔ Answer: x = 6

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Problem 8

Given: Right angle symbol at Y — so ∠ZYH = 90°, and →YZ bisects ∠TYH
Wait — actually, looking at the diagram, →YZ is the bisector of ∠TYH, and there’s a right angle between ZY and HY? Actually, the right angle is at Y between lines TY and HY? Let me re-read.

Actually, the diagram shows →YZ bisecting ∠TYH, and the angle between ZY and HY is marked with a square — meaning ∠ZYH = 90°? But if YZ is the bisector, then ∠TYZ = ∠ZYH.

Wait — no, if YZ is the bisector, then ∠TYZ = ∠ZYH. But here, ∠ZYH is labeled as 90°? That would mean ∠TYZ is also 90°, so total angle ∠TYH = 180° — which is possible.

But the expression given is for ∠TYZ = 4x + 74°, and since it equals ∠ZYH = 90°:

So:
4x + 74 = 90
→ 4x = 16
→ x = 4

✔ Answer: x = 4

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Problem 9

Given: ∠UYZ = 15°, ∠DYZ = 2x - 3°, and →YZ bisects ∠UYD
So, 2x - 3 = 15
→ 2x = 18
→ x = 9

✔ Answer: x = 9

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Problem 10

Given: ∠EYZ = 3x°, ∠ZYQ = x + 30°, and →YZ bisects ∠EYQ
So, 3x = x + 30
→ 2x = 30
→ x = 15

✔ Answer: x = 15

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Problem 11

Given: ∠AYZ = 10 + 2x, ∠LYZ = 4x, and →YZ bisects ∠AYL
So, 10 + 2x = 4x
→ 10 = 2x
→ x = 5

✔ Answer: x = 5

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Problem 12

Given: ∠ZYG = 4x, ∠SYZ = 3x + 140°, and →YZ bisects ∠SYG
So, 4x = 3x + 140
→ x = 140

✔ Answer: x = 140

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Problem 13

Given: ∠TYZ = x + 60°, ∠HYZ = 3x°, and →YZ bisects ∠TYH
So, x + 60 = 3x
→ 60 = 2x
→ x = 30

✔ Answer: x = 30

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Problem 14

Given: ∠DYZ = 4x + 14°, ∠UYZ = 6x°, and →YZ bisects ∠DYU
So, 4x + 14 = 6x
→ 14 = 2x
→ x = 7

✔ Answer: x = 7

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Problem 15

Given: ∠EYZ = 9x°, and →YZ bisects ∠EYQ
But we are not given the other angle directly. However, since →YZ is the bisector, and assuming the diagram implies ∠EYQ is split into two equal parts, and one part is 9x, then the other part is also 9x.

Wait — but we need to find x? There’s no other value given. Unless... maybe the total angle is implied? Or perhaps we’re to assume something?

Looking back — in Problem 5, we had 88°, here we have 9x, and no other number. But if YZ is the bisector, then both angles are equal — so unless we’re told the total or another value, we can’t solve numerically.

Wait — perhaps this is a trick? Or maybe the diagram implies that ∠EYQ is a straight angle? No, EY is vertical, YQ is horizontal — so ∠EYQ is 90°!

Ah! Look: Point E is above Y, Q is to the right — so ∠EYQ is a right angle, 90°.

Since →YZ bisects it, each half is 45°.

So, 9x = 45
→ x = 5

✔ Answer: x = 5

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## ✔ Final Answers:

1) 36°
2) 78°
3) 108°
4) 70°
5) 176°
6) x = 3
7) x = 6
8) x = 4
9) x = 9
10) x = 15
11) x = 5
12) x = 140
13) x = 30
14) x = 7
15) x = 5

All solved using the definition: Angle bisector creates two equal angles.