Let's consider the following diagram.

We want to determine $E F .$ To do so we will use the fact that all pairs of corresponding sides of two similar figures are proportional. First let's find $A E$ and $E C .$

We can see that

∠ D E A

and

∠ C E B

are vertical angles, so

∠ D E A ≅ ∠ C E B .

Since

\overline{A D} ∥ \overline{B C},

∠ A D E ≅ ∠ E B C

by the Vertical Angles Congruence Theorem. Two angles of

△ A E D

are congruent to two angles of

△ C E B,

so

△ A E D

and

△ C E B

are similar by the Angle-Angle (AA) Similarity Theorem.

△ A E D \sim △ C E B

Note that

\overline{A D}

corresponds to

\overline{B C},

and

\overline{A E}

corresponds to

\overline{E C} .

Recall that all pairs of corresponding sides of two similar figures are proportional.

\frac{A D}{B C} = \frac{A E}{E C}

We know that

A D = 40

and

B C = 30 .

Let's substitute these values into the above equation and determine

A E

in terms of

E C .

\frac{A D}{B C} = \frac{A E}{E C}

SubstituteII

$A D = 40$ , $B C = 30$

\frac{4 0}{3 0} = \frac{A E}{E C}

▼

Solve for

A E

RearrangeEqn

Rearrange equation

\frac{A E}{E C} = \frac{4 0}{3 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 0}{b / 1 0}$

\frac{A E}{E C} = \frac{4}{3}

MultEqn

$LHS \cdot E C = RHS \cdot E C$

A E = \frac{4}{3} E C

We have that

A E

is equal to four-thirds of

E C .

Let's also find

A C

in terms of

E C .

A C = A E + E C

Substitute

$A E = \frac{4}{3} E C$

A C = \frac{4}{3} E C + E C

▼

Add terms

Rewrite

Rewrite $E C$ as $1 \cdot E C$

A C = \frac{4}{3} E C + 1 \cdot E C

OneToFrac

Rewrite $1$ as $\frac{3}{3}$

A C = \frac{4}{3} E C + \frac{3}{3} E C

AddFrac

Add fractions

A C = \frac{7}{3} E C

We have that

A C

is equal to seven-thirds of

E C .

Now we will move on to determining

E F .

First we have to find a triangle similar to

△ F C E .

Angles

∠ E F C

and

∠ A D C

are both right angles, so they are congruent. Triangles

△ F C E

and

△ D C A

share an angle,

∠ F C E .

In conclusion, we have that two angles of

△ F C E

are congruent to two angles of

△ D C A,

so

△ F C E

is similar to

△ D C A

by the AA Similarity Theorem.

△ D C A \sim △ F C E

Note that

\overline{E F}

corresponds to

\overline{A D},

and

\overline{E C}

corresponds to

\overline{A C} .

Remember that

A D = 40

and

A C = \frac{7}{3} E C .

\frac{E F}{A D} = \frac{E C}{A C} \Leftrightarrow \frac{E F}{4 0} = \frac{E C}{7 / 3 E C}

Let's solve the obtained equation for

E F .

\frac{E F}{4 0} = \frac{E C}{7 / 3 E C}

▼

Solve for

E F

ReduceFrac

$\frac{a}{b} = \frac{a / E C}{b / E C}$

\frac{E F}{4 0} = \frac{1}{7 / 3}

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

\frac{E F}{4 0} = \frac{3}{7}

MultEqn

$LHS \cdot 40 = RHS \cdot 40$

E F = \frac{1 2 0}{7}

UseCalc

Use a calculator

E F \approx 17.1

We have that

E F

is about

17.1

feet.

Exercise