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### **Problem 17**
**If \( K \) is the incenter of \( \triangle EFG \):**
- \( K \) is the point where the angle bisectors meet.
- The angle measures at \( \triangle EFG \) must add up to \( 180^\circ \).

We have the given angles:
1. \( \angle KEJ = 18x - 23 \)
2. \( \angle JEG = 7x - 1 \)
3. \( \angle EFG = 47^\circ \).

**Solution:**
The sum of all angles in \( \triangle EFG \) is:
\[
(18x - 23) + (7x - 1) + 47 = 180
\]

Simplify:
\[
25x + 23 = 180
\]

Solve for \( x \):
\[
25x = 157 \quad \implies \quad x = 6.28 \, \text{(approximately)}.
\]

**Now, find each angle measure:**
1. \( \angle KEJ = 18x - 23 = 18(6.28) - 23 = 113.04 - 23 = 90.04^\circ \).
2. \( \angle JEG = 7x - 1 = 7(6.28) - 1 = 43.96^\circ \).
3. \( \angle EFG = 47^\circ \).
4. \( \angle FGE = 180 - (\angle KEJ + \angle JEG) = 180 - (90.04 + 43.96) = 46^\circ \).

---

### **Problem 18**
**If \( P \) is the centroid of \( \triangle JKL \):**
- A centroid divides each median into a \( 2:1 \) ratio.

Given:
1. \( PN = 5 \),
2. \( LM = 33 \),
3. \( KP = 16 \),
4. \( NL = 13 \).

**Solution:**
1. **\( PM \):** Since \( P \) divides \( LM \) into a \( 2:1 \) ratio:
\[
PM = \frac{2}{3} \times 33 = 22
\]

2. **\( PL \):** Since \( P \) divides \( KL \) into a \( 2:1 \) ratio:
\[
PL = \frac{2}{3} \times KP = \frac{2}{3} \times 16 = 10.67
\]

3. **\( PO \):** Since \( P \) divides \( NL \) into a \( 2:1 \) ratio:
\[
PO = \frac{2}{3} \times 13 = 8.67
\]

4. **\( JP \):** From \( JP = PN + PO \):
\[
JP = 5 + 8.67 = 13.67
\]

5. **\( JN \):** Sum \( JP + PN \):
\[
JN = 13.67 + 5 = 18.67
\]

6. **\( KL \):** Total segment \( KL = 3 \times KP / 2 \):
\[
KL = 3 \times 16 / 2 = 24
\]

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### Final Answers:

**Problem 17:**
- \( x = 6.28 \)
- \( m\angle KEJ = 90.04^\circ \)
- \( m\angle EFG = 47^\circ \)
- \( m\angle FGE = 46^\circ \)
- \( m\angle KGJ = \) (can be calculated based on geometry, assuming symmetry).

**Problem 18:**
- \( PM = 22 \)
- \( PL = 10.67 \)
- \( PO = 8.67 \)
- \( JP = 13.67 \)
- \( JN = 18.67 \)
- \( KL = 24 \)

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Do you need further clarification or detailed explanations for any part of this solution? 

Here are 5 additional questions to expand your understanding:
1. How can the centroid formula be derived geometrically?
2. What are the properties of an incenter in a triangle?
3. Why does the sum of angles in a triangle always equal \( 180^\circ \)?
4. How do you confirm the \( 2:1 \) ratio property of the centroid?
5. How can angle bisectors help in solving triangle properties?

**Tip:** Always double-check your calculations, especially when working with angle measures and triangle properties!

Solve the geometry problems: 17. If K is the incenter of ΔEFG, find x and each angle measure. 18. If P is the centroid of ΔJKL, find each measure given segment details.

Learn how to solve geometry problems involving the incenter and centroid of triangles. Problem 17 explores finding x and angle measures when K is the incenter of ΔEFG, while Problem 18 deals with centroid P dividing medians in a 2:1 ratio in ΔJKL. Includes step-by-step solutions for students in Grades 9-11.

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