Geometry Review Worksheet featuring problems on similar triangles, parallel lines, and parallelograms.

Problem 1: Given \(\overline{DE} \parallel \overline{BC}\)

When \(\overline{DE} \parallel \overline{BC}\), the 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. We will use this property to solve each part.

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

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

Since \(\triangle ADE \sim \triangle ABC\), the ratio of the corresponding sides is:
\[
\frac{AD}{AB} = \frac{DE}{BC}
\]

Substitute the known values:
\[
\frac{7}{10} = \frac{6}{BC}
\]

Solve for \(BC\):
\[
7 \cdot BC = 6 \cdot 10 \implies 7 \cdot BC = 60 \implies BC = \frac{60}{7}
\]

Thus, the answer is:
\[
\boxed{\frac{60}{7}}
\]

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

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

Since \(\triangle ADE \sim \triangle ABC\), the ratio of the corresponding sides is:
\[
\frac{AD}{AB} = \frac{AE}{AC}
\]

Substitute the known values:
\[
\frac{3}{8} = \frac{4}{AC}
\]

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

Now, find \(CE\):
\[
CE = AC - AE = \frac{32}{3} - 4 = \frac{32}{3} - \frac{12}{3} = \frac{20}{3}
\]

Thus, the answer is:
\[
\boxed{\frac{20}{3}}
\]

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

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

Since \(\triangle ADE \sim \triangle ABC\), the ratio of the corresponding sides is:
\[
\frac{AD}{AB} = \frac{DE}{BC}
\]

Substitute the known values:
\[
\frac{4}{10} = \frac{DE}{25}
\]

Simplify the fraction:
\[
\frac{2}{5} = \frac{DE}{25}
\]

Solve for \(DE\):
\[
2 \cdot 25 = 5 \cdot DE \implies 50 = 5 \cdot DE \implies DE = 10
\]

Thus, the answer is:
\[
\boxed{10}
\]

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

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

Since \(\triangle ADE \sim \triangle ABC\), the ratio of the corresponding sides is:
\[
\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}
\]

Using \(\frac{AD}{AB} = \frac{AE}{AC}\):
\[
\frac{x-1}{x+4} = \frac{1}{AC}
\]

Also, \(AC = AE + CE = 1 + (x+3) = x + 4\). So:
\[
\frac{x-1}{x+4} = \frac{1}{x+4}
\]

This equation is always true, so we use the other ratio \(\frac{DE}{BC}\):
\[
\frac{AD}{AB} = \frac{DE}{BC} \implies \frac{x-1}{x+4} = \frac{2x+1}{BC}
\]

We already know \(AC = x + 4\), so:
\[
BC = AC \cdot \frac{DE}{AE} = (x+4) \cdot \frac{2x+1}{1} = (x+4)(2x+1)
\]

Using the proportion:
\[
\frac{x-1}{x+4} = \frac{2x+1}{(x+4)(2x+1)/(x+4)} = \frac{2x+1}{2x+1} = 1
\]

This is consistent. Now solve for \(x\) using the proportion:
\[
\frac{x-1}{x+4} = \frac{2x+1}{25}
\]

Cross-multiply:
\[
(x-1) \cdot 25 = (2x+1) \cdot (x+4)
\]
\[
25x - 25 = 2x^2 + 8x + x + 4
\]
\[
25x - 25 = 2x^2 + 9x + 4
\]
\[
0 = 2x^2 + 9x + 4 - 25x + 25
\]
\[
0 = 2x^2 - 16x + 29
\]

Solve the quadratic equation:
\[
x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 2 \cdot 29}}{2 \cdot 2} = \frac{16 \pm \sqrt{256 - 232}}{4} = \frac{16 \pm \sqrt{24}}{4} = \frac{16 \pm 2\sqrt{6}}{4} = \frac{8 \pm \sqrt{6}}{2}
\]

Since \(x\) must be a positive value, we take:
\[
x = \frac{8 - \sqrt{6}}{2}
\]

Now find \(BC\):
\[
BC = 25
\]

Thus, the answers are:
\[
\boxed{4, 25}
\]

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

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

Since \(\triangle ADE \sim \triangle ABC\), the ratio of the corresponding sides is:
\[
\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}
\]

Using \(\frac{AD}{AB} = \frac{AE}{AC}\):
\[
\frac{2x}{3x+3} = \frac{4x-1}{AC}
\]

Also, \(AC = AE + CE = (4x-1) + 5x = 9x - 1\). So:
\[
\frac{2x}{3x+3} = \frac{4x-1}{9x-1}
\]

Cross-multiply:
\[
2x(9x-1) = (4x-1)(3x+3)
\]
\[
18x^2 - 2x = 12x^2 + 12x - 3x - 3
\]
\[
18x^2 - 2x = 12x^2 + 9x - 3
\]
\[
6x^2 - 11x + 3 = 0
\]

Solve the quadratic equation:
\[
x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4 \cdot 6 \cdot 3}}{2 \cdot 6} = \frac{11 \pm \sqrt{121 - 72}}{12} = \frac{11 \pm \sqrt{49}}{12} = \frac{11 \pm 7}{12}
\]

So:
\[
x = \frac{18}{12} = \frac{3}{2} \quad \text{or} \quad x = \frac{4}{12} = \frac{1}{3}
\]

Since \(x = \frac{3}{2}\) is the valid solution, we find \(DE\):
\[
DE = \frac{AD}{AB} \cdot BC = \frac{2x}{3x+3} \cdot (6x+2) = \frac{2 \cdot \frac{3}{2}}{3 \cdot \frac{3}{2} + 3} \cdot \left(6 \cdot \frac{3}{2} + 2\right) = \frac{3}{\frac{9}{2} + 3} \cdot (9 + 2) = \frac{3}{\frac{15}{2}} \cdot 11 = \frac{3 \cdot 2}{15} \cdot 11 = \frac{6}{15} \cdot 11 = \frac{2}{5} \cdot 11 = \frac{22}{5}
\]

Thus, the answers are:
\[
\boxed{\frac{3}{2}, \frac{22}{5}}
\]

Problem 2: Given \(\angle 1 \cong \angle 2\)

When \(\angle 1 \cong \angle 2\), the triangles are similar by the AA similarity criterion. We will use the property of similar triangles to solve each part.

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

Since \(\triangle ACD \sim \triangle BCA\), the ratio of the corresponding sides is:
\[
\frac{AD}{AC} = \frac{AC}{BC}
\]

Substitute the known values:
\[
\frac{AD}{6} = \frac{6}{8} = \frac{3}{4}
\]

Solve for \(AD\):
\[
AD = 6 \cdot \frac{3}{4} = \frac{18}{4} = \frac{9}{2}
\]

Thus, the answer is:
\[
\boxed{\frac{9}{2}}
\]

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

Since \(\triangle ACD \sim \triangle BCA\), the ratio of the corresponding sides is:
\[
\frac{AD}{AC} = \frac{AC}{BC}
\]

Substitute the known values:
\[
\frac{AD}{4} = \frac{4}{8} = \frac{1}{2}
\]

Solve for \(AD\):
\[
AD = 4 \cdot \frac{1}{2} = 2
\]

Thus, the answer is:
\[
\boxed{2}
\]

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

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

Since \(\triangle ACD \sim \triangle BCA\), the ratio of the corresponding sides is:
\[
\frac{AD}{AC} = \frac{AC}{BC}
\]

Substitute the known values:
\[
\frac{x-4}{3} = \frac{3}{x}
\]

Cross-multiply:
\[
(x-4)x = 3 \cdot 3
\]
\[
x^2 - 4x = 9
\]
\[
x^2 - 4x - 9 = 0
\]

Solve the quadratic equation:
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 36}}{2} = \frac{4 \pm \sqrt{52}}{2} = \frac{4 \pm 2\sqrt{13}}{2} = 2 \pm \sqrt{13}
\]

Since \(x\) must be positive, we take:
\[
x = 2 + \sqrt{13}
\]

Thus, the answer is:
\[
\boxed{2 + \sqrt{13}}
\]

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

In a parallelogram, opposite sides are equal, and the diagonals bisect each other. We will use the properties of similar triangles to solve for the unknowns.

#### Find: \(BE\), \(CE\), \(CF\).

Since \(ABCD\) is a parallelogram, \(AB = CD = 12\) and \(AD = BC = 8\). The diagonal \(AC\) intersects \(DE\) at point \(E\). By the properties of similar triangles, we have:
\[
\frac{BE}{EC} = \frac{AB}{DC} = 1
\]

So, \(BE = EC\). Let \(BE = x\). Then \(EC = x\) and \(BC = BE + EC = x + x = 2x = 8\). Thus:
\[
x = 4
\]

So, \(BE = 4\) and \(CE = 4\).

Now, consider \(\triangle ABE \sim \triangle CDF\). The ratio of the corresponding sides is:
\[
\frac{AB}{CD} = \frac{AE}{CF}
\]

Since \(AB = CD = 12\), the ratio is 1. So:
\[
AE = CF
\]

Given \(AE = 10\) and \(EF = 4\), we have:
\[
AF = AE + EF = 10 + 4 = 14
\]

Since \(ABCD\) is a parallelogram, \(AD = BC = 8\). So:
\[
CF = AF - AC = 14 - 8 = 6
\]

Thus, the answers are:
\[
\boxed{4, 4, 6}
\]

Problem 4: Given \(l \parallel m \parallel n\)

When lines are parallel, the segments intercepted by transversals are proportional. We will use this property to solve for \(x\).

#### Find: \(x\).

The segments intercepted by the transversals are proportional:
\[
\frac{3}{5} = \frac{x}{x+2}
\]

Cross-multiply:
\[
3(x+2) = 5x
\]
\[
3x + 6 = 5x
\]
\[
6 = 2x
\]
\[
x = 3
\]

Thus, the answer is:
\[
\boxed{3}
\]