Worksheet 1: Geometry problem involving a perpendicular bisector and angle calculations in a triangle.

It looks like you've uploaded a worksheet with geometry problems, but the image is slightly cut off and the text in question 5 is incomplete (especially part a). However, based on what's visible — particularly the triangle diagram with labels and expressions involving $ y $ — I can help solve question 5, which appears to be the main focus.

Let’s go through Question 5 step by step, using the information from the diagram:

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🔷 Given Diagram:

- Triangle $ \triangle EHF $
- Point $ G $ lies on segment $ HF $
- Segment $ EG $ is perpendicular to $ HF $, and it bisects $ HF $ (since there are tick marks showing $ HG = GF $)
- Angles at point $ E $:
- $ \angle GEH = 2y + 18^\circ $
- $ \angle GEF = 4y^\circ $
- $ EG $ is drawn as a perpendicular line from $ E $ to $ HF $, meeting at $ G $

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## ✔ Problem 5:

a) ________ is the perpendicular bisector of segment ________.

From the diagram:
- $ EG $ is perpendicular to $ HF $, and it divides $ HF $ into two equal parts ($ HG = GF $).
- So, $ EG $ is the perpendicular bisector of segment $ HF $.

✔ Answer:
> EG is the perpendicular bisector of segment HF.

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b) What are the lengths of $ \overline{EF} $ and $ \overline{EH} $?

We don't have numerical lengths given directly, but we can infer symmetry.

Since $ EG $ is the perpendicular bisector of $ HF $, and $ E $ is a vertex above the base $ HF $, then:

- $ EH = EF $ because any point on the perpendicular bisector of a segment is equidistant from the endpoints.
- So $ \triangle EHF $ is isosceles with $ EH = EF $

But no actual length values are given in the diagram or problem. So unless there's a missing piece of data (like a scale or side label), we cannot compute exact lengths.

⚠️ Possibility: Maybe the tick marks on $ EH $ and $ EF $ suggest they are congruent? Yes — there are tick marks on both $ EH $ and $ EF $, indicating they are congruent.

So, without specific numbers, we can only say:

✔ Answer:
> $ EH = EF $, but their exact lengths cannot be determined from the given information.

*(If this was meant to be solved numerically, perhaps a length was supposed to be labeled. But based on current info, we can only state equality.)*

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c) Find the value of $ y $.

We are told:
- $ m\angle GEH = 2y + 18^\circ $
- $ m\angle GEF = 4y^\circ $
- Since $ EG $ is the perpendicular bisector, and $ \angle HEF $ is split into two angles at $ E $, and since $ EG $ is perpendicular to $ HF $, the total angle $ \angle HEF $ is composed of $ \angle GEH $ and $ \angle GEF $, and together they form the full angle at $ E $.

But more importantly, notice that $ EG $ is perpendicular to $ HF $, so $ \angle EGH = \angle EGF = 90^\circ $. This means that $ \angle HEF $ is split into two parts by $ EG $, and since $ EG $ is the angle bisector of $ \angle HEF $ only if the triangle is symmetric — which it is (because $ EH = EF $), so yes, $ EG $ is also the angle bisector.

Wait — actually, we don’t need to assume that yet.

Let’s use the fact that $ \angle GEH $ and $ \angle GEF $ are adjacent angles forming $ \angle HEF $, and since $ EG $ is perpendicular to $ HF $, the angles at $ G $ are right angles.

But here’s a better idea: since $ EG $ is the perpendicular bisector, and $ EH = EF $, then $ \triangle EHG \cong \triangle EFG $ by SAS (or HL), so corresponding angles are equal.

But let’s look at the angles at point E.

We know:
- $ \angle GEH = 2y + 18^\circ $
- $ \angle GEF = 4y^\circ $
- These two angles make up $ \angle HEF $

But there’s no direct sum given. However, notice that in the diagram, these two angles appear to be adjacent, and together they form the angle at $ E $. But unless we know $ \angle HEF $, we can’t add them.

Wait — but here’s a key insight: Since $ EG $ is perpendicular to $ HF $, and $ G $ is the midpoint, then $ \triangle EHG $ and $ \triangle EFG $ are both right triangles, and since $ HG = GF $, and $ EG $ is common, and $ EH = EF $, then $ \triangle EHG \cong \triangle EFG $ by HL or SAS.

Therefore, the angles at $ E $ must be equal:
> $ \angle GEH = \angle GEF $

So:
$$
2y + 18 = 4y
$$

Now solve for $ y $:

$$
2y + 18 = 4y \\
18 = 4y - 2y \\
18 = 2y \\
y = 9
$$

✔ Answer: $ y = 9 $

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d) Find $ m\angle GEH $ and $ m\angle GEF $.

Now plug $ y = 9 $ into the expressions:

- $ m\angle GEH = 2y + 18 = 2(9) + 18 = 18 + 18 = 36^\circ $
- $ m\angle GEF = 4y = 4(9) = 36^\circ $

✔ Answer:
> $ m\angle GEH = 36^\circ $, $ m\angle GEF = 36^\circ $

This makes sense — they are equal, as expected due to symmetry.

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✔ Final Answers:

#### 5a)
> $ EG $ is the perpendicular bisector of segment $ HF $.

#### 5b)
> $ EH = EF $, but exact lengths not provided. (Assuming no measurements were given.)

#### 5c)
> $ y = 9 $

#### 5d)
> $ m\angle GEH = 36^\circ $, $ m\angle GEF = 36^\circ $

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🟦 Bonus: Questions 6 & 7

These are construction tasks:

#### 6. a) Draw the angle bisector of $ \angle A $
- Use a compass to draw arcs from point $ A $ intersecting sides $ AB $ and $ AC $
- From those points, draw arcs inside the triangle that intersect
- Connect $ A $ to that intersection point — that’s the angle bisector

b) Mark congruent angles:
- Put one arc on each side of $ \angle A $, showing that the two resulting angles are equal.

#### 7. a) Draw the perpendicular bisector of $ DF $
- Find the midpoint of $ DF $
- Construct a line perpendicular to $ DF $ passing through the midpoint

b) Mark right angles and congruent segments:
- Put a small square at the intersection to show right angles
- Add tick marks on $ DG $ and $ GF $ to show they are equal

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Let me know if you'd like a visual explanation or sketch description!