Geometry Review Worksheet with problems on similar triangles, parallel lines, parallelograms, and angle relationships.

Problem Analysis and Solutions

The worksheet involves several geometry problems related to similar triangles, properties of parallelograms, and parallel lines. Let's solve each part step by step.

---

#### (1) Refer to the figure to the right, given \(\overline{DE} \parallel \overline{BC}\).

Given that \(\overline{DE} \parallel \overline{BC}\), triangles \( \triangle ADE \) and \( \triangle ABC \) are similar by the Basic Proportionality Theorem (also known as Thales' Theorem). This means the corresponding sides are proportional:
\[
\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}
\]

##### (a) AD = 7, BD = 3, DE = 6. Find: BC.

1. First, find \( AB \):
\[
AB = AD + BD = 7 + 3 = 10
\]

2. Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{DE}{BC}
\]
Substitute the known values:
\[
\frac{7}{10} = \frac{6}{BC}
\]

3. Solve for \( BC \):
\[
BC = \frac{6 \times 10}{7} = \frac{60}{7}
\]

Answer:
\[
\boxed{\frac{60}{7}}
\]

##### (b) AD = 3, BD = 5, AE = 4. Find: CE.

1. First, find \( AB \):
\[
AB = AD + BD = 3 + 5 = 8
\]

2. Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{AE}{AC}
\]
Substitute the known values:
\[
\frac{3}{8} = \frac{4}{AC}
\]

3. Solve for \( AC \):
\[
AC = \frac{4 \times 8}{3} = \frac{32}{3}
\]

4. Since \( AC = AE + CE \):
\[
CE = AC - AE = \frac{32}{3} - 4 = \frac{32}{3} - \frac{12}{3} = \frac{20}{3}
\]

Answer:
\[
\boxed{\frac{20}{3}}
\]

##### (c) AD = 4, AB = 10, BC = 25. Find: DE.

1. First, find \( BD \):
\[
BD = AB - AD = 10 - 4 = 6
\]

2. Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{DE}{BC}
\]
Substitute the known values:
\[
\frac{4}{10} = \frac{DE}{25}
\]

3. Solve for \( DE \):
\[
DE = \frac{4 \times 25}{10} = 10
\]

Answer:
\[
\boxed{10}
\]

##### (d) AD = (x - 1), BD = 5, AE = 1, CE = (x + 3), DE = (2x + 1). Find: x, BC.

1. First, find \( AB \):
\[
AB = AD + BD = (x - 1) + 5 = x + 4
\]

2. Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{AE}{AC}
\]
Substitute the known values:
\[
\frac{x - 1}{x + 4} = \frac{1}{1 + (x + 3)} = \frac{1}{x + 4}
\]

3. Solve for \( x \):
\[
\frac{x - 1}{x + 4} = \frac{1}{x + 4}
\]
Cross-multiply:
\[
(x - 1)(x + 4) = 1(x + 4)
\]
Simplify:
\[
x^2 + 4x - x - 4 = x + 4
\]
\[
x^2 + 3x - 4 = x + 4
\]
\[
x^2 + 2x - 8 = 0
\]
Factorize:
\[
(x + 4)(x - 2) = 0
\]
So, \( x = -4 \) or \( x = 2 \). Since \( x \) must be positive:
\[
x = 2
\]

4. Find \( BC \):
Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{DE}{BC}
\]
Substitute \( x = 2 \):
\[
AD = x - 1 = 2 - 1 = 1, \quad AB = x + 4 = 2 + 4 = 6, \quad DE = 2x + 1 = 2(2) + 1 = 5
\]
\[
\frac{1}{6} = \frac{5}{BC}
\]
Solve for \( BC \):
\[
BC = 5 \times 6 = 30
\]

Answer:
\[
\boxed{2, 30}
\]

##### (e) AD = 2x, BD = (x + 3), AE = (4x - 1), CE = 5x, BC = (6x + 2). Find: x, DE.

1. First, find \( AB \):
\[
AB = AD + BD = 2x + (x + 3) = 3x + 3
\]

2. Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{AE}{AC}
\]
Substitute the known values:
\[
\frac{2x}{3x + 3} = \frac{4x - 1}{4x - 1 + 5x} = \frac{4x - 1}{9x - 1}
\]

3. Solve for \( x \):
\[
\frac{2x}{3x + 3} = \frac{4x - 1}{9x - 1}
\]
Cross-multiply:
\[
2x(9x - 1) = (4x - 1)(3x + 3)
\]
Expand both sides:
\[
18x^2 - 2x = 12x^2 + 12x - 3x - 3
\]
\[
18x^2 - 2x = 12x^2 + 9x - 3
\]
Simplify:
\[
6x^2 - 11x + 3 = 0
\]
Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 6, \, b = -11, \, c = 3
\]
\[
x = \frac{11 \pm \sqrt{121 - 72}}{12} = \frac{11 \pm \sqrt{49}}{12} = \frac{11 \pm 7}{12}
\]
So, \( x = \frac{18}{12} = \frac{3}{2} \) or \( x = \frac{4}{12} = \frac{1}{3} \). Since \( x = \frac{1}{3} \) does not satisfy the problem constraints, we take:
\[
x = \frac{3}{2}
\]

4. Find \( DE \):
Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{DE}{BC}
\]
Substitute \( x = \frac{3}{2} \):
\[
AD = 2x = 2 \left( \frac{3}{2} \right) = 3, \quad AB = 3x + 3 = 3 \left( \frac{3}{2} \right) + 3 = \frac{9}{2} + 3 = \frac{15}{2}, \quad BC = 6x + 2 = 6 \left( \frac{3}{2} \right) + 2 = 9 + 2 = 11
\]
\[
\frac{3}{\frac{15}{2}} = \frac{DE}{11}
\]
Simplify:
\[
\frac{3 \times 2}{15} = \frac{DE}{11}
\]
\[
\frac{6}{15} = \frac{DE}{11}
\]
\[
\frac{2}{5} = \frac{DE}{11}
\]
Solve for \( DE \):
\[
DE = \frac{2 \times 11}{5} = \frac{22}{5}
\]

Answer:
\[
\boxed{\frac{3}{2}, \frac{22}{5}}
\]

---

#### (2) Refer to the figure to the right, \(\angle 1 \cong \angle 2\).

Given \(\angle 1 \cong \angle 2\), triangles \( \triangle ACD \) and \( \triangle ABC \) are similar by the AA (Angle-Angle) similarity criterion. This means the corresponding sides are proportional:
\[
\frac{AD}{AB} = \frac{AC}{AC + CB}
\]

##### (a) AC = 6, BC = 8, BD = 5. Find: AD.

1. First, find \( AB \):
\[
AB = AD + BD
\]

2. Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{AC}{AC + CB}
\]
Substitute the known values:
\[
\frac{AD}{AD + 5} = \frac{6}{6 + 8} = \frac{6}{14} = \frac{3}{7}
\]

3. Solve for \( AD \):
\[
\frac{AD}{AD + 5} = \frac{3}{7}
\]
Cross-multiply:
\[
7AD = 3(AD + 5)
\]
Simplify:
\[
7AD = 3AD + 15
\]
\[
4AD = 15
\]
\[
AD = \frac{15}{4}
\]

Answer:
\[
\boxed{\frac{15}{4}}
\]

##### (b) AB = 10, AC = 4, BC = 8. Find: AD.

1. Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{AC}{AC + CB}
\]
Substitute the known values:
\[
\frac{AD}{10} = \frac{4}{4 + 8} = \frac{4}{12} = \frac{1}{3}
\]

2. Solve for \( AD \):
\[
\frac{AD}{10} = \frac{1}{3}
\]
\[
AD = \frac{10}{3}
\]

Answer:
\[
\boxed{\frac{10}{3}}
\]

##### (c) AC = 3, AD = (x - 4), BC = x, BD = 4. Find: BC.

1. First, find \( AB \):
\[
AB = AD + BD = (x - 4) + 4 = x
\]

2. Use the similarity ratio:
\[
\frac{AD}{AB} = \frac{AC}{AC + CB}
\]
Substitute the known values:
\[
\frac{x - 4}{x} = \frac{3}{3 + x}
\]

3. Solve for \( x \):
\[
\frac{x - 4}{x} = \frac{3}{3 + x}
\]
Cross-multiply:
\[
(x - 4)(3 + x) = 3x
\]
Expand both sides:
\[
3x + x^2 - 12 - 4x = 3x
\]
Simplify:
\[
x^2 - x - 12 = 0
\]
Factorize:
\[
(x - 4)(x + 3) = 0
\]
So, \( x = 4 \) or \( x = -3 \). Since \( x \) must be positive:
\[
x = 4
\]

Answer:
\[
\boxed{4}
\]

---

#### (3) Given: \( ABCD \) is a parallelogram, sides as marked.

In a parallelogram, opposite sides are equal, and diagonals bisect each other. We need to find \( BE \), \( CE \), and \( CF \).

1. Since \( ABCD \) is a parallelogram:
\[
AB = CD = 12, \quad AD = BC = 8
\]

2. Diagonal \( AC \) bisects \( BD \) at \( E \). By the properties of parallelograms:
\[
BE = ED, \quad AE = EC
\]

3. Use the triangle similarity in \( \triangle ABE \) and \( \triangle CDE \):
\[
\frac{BE}{ED} = \frac{AB}{CD} = 1 \quad \text{(since \( AB = CD \))}
\]
So, \( BE = ED \).

4. Use the given side lengths to find the segments:
\[
BE = \frac{BD}{2}, \quad CE = \frac{AC}{2}
\]
Since \( BD = 10 \):
\[
BE = ED = \frac{10}{2} = 5
\]
Since \( AC = 16 \) (by the Pythagorean theorem in \( \triangle ACD \)):
\[
CE = \frac{16}{2} = 8
\]
Finally, \( CF \) is the extension of \( CE \) such that \( CF = CE + EF \). Since \( EF = 4 \):
\[
CF = 8 + 4 = 12
\]

Answer:
\[
\boxed{5, 8, 12}
\]

---

#### (4) Given: The figure below, \( l \parallel m \parallel n \).

Given parallel lines, the segments cut by the transversals are proportional. We need to find \( x \).

1. Use the proportionality of the segments:
\[
\frac{3}{5} = \frac{x}{x + 2}
\]

2. Solve for \( x \):
\[
\frac{3}{5} = \frac{x}{x + 2}
\]
Cross-multiply:
\[
3(x + 2) = 5x
\]
Simplify:
\[
3x + 6 = 5x
\]
\[
6 = 2x
\]
\[
x = 3
\]

Answer:
\[
\boxed{3}
\]

---

Final Answers:

1. (a) \(\boxed{\frac{60}{7}}\)
(b) \(\boxed{\frac{20}{3}}\)
(c) \(\boxed{10}\)
(d) \(\boxed{2, 30}\)
(e) \(\boxed{\frac{3}{2}, \frac{22}{5}}\)

2. (a) \(\boxed{\frac{15}{4}}\)
(b) \(\boxed{\frac{10}{3}}\)
(c) \(\boxed{4}\)

3. \(\boxed{5, 8, 12}\)

4. \(\boxed{3}\)