Let's begin by sketching the triangle with its angle bisectors.

Considering the Triangle-Angle-Bisector Theorem, let's find the lengths of each pair of segments one at a time.

Lengths of $\overline{B D}$ and $\overline{D C}$

We will first consider $B D$ and $D C .$

By the theorem, we can write a proportion for

B D

and

D C .

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

We have already been given

A B = 5

and

A C = 12 .

In this case, to find

B D

and

D C,

we should identify a relation between them. Knowing that

B C = 13

and considering the Segment Addition Postulate, let's write

D C

in terms of

B D .

B C = B D + D C ⇓ D C = 13 - B D

Next, we will substitute the given values and

D C = 13 - B D

into the proportion, so we can find the length of

\overline{B D} .

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

SubstituteValues

Substitute values

\frac{5}{1 2} = \frac{B D}{1 3 - B D}

▼

Solve for

B D

CrossMult

Cross multiply

5 (13 - B D) = 12 B D

Distr

Distribute $5$

65 - 5 B D = 12 B D

AddEqn

$LHS + 5 B D = RHS + 5 B D$

65 = 17 B D

DivEqn

$LHS / 17 = RHS / 17$

\frac{6 5}{1 7} = B D

RearrangeEqn

Rearrange equation

B D = \frac{6 5}{1 7}

CalcQuot

Calculate quotient

B D = 3.82352 \dots

RoundDec

Round to $1$ decimal place(s)

B D \approx 3.8

Now, we can find

D C

by substituting

B D \approx 3.8

into

D C = 13 - B D .

D C = 13 - B D

Substitute

$B D = 3.8$

D C = 13 - 3.8

SubTerm

Subtract term

D C \approx 9.2

Lengths of $\overline{A E}$ and $\overline{E C}$

As our second pair of segments, we will think of $\overline{A E}$ and $\overline{E C} .$

Considering the Triangle-Angle-Bisector Theorem, let's write a proportion to find

A E

and

E C .

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

Next, we will rewrite

E C

in terms of

A E

using the fact that

A C = 12 .

A C = A E + E C ⇓ E C = 12 - A E

Now, we will first find

A E

by substituting the value into the proportion.

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

SubstituteValues

Substitute values

\frac{5}{1 3} = \frac{A E}{1 2 - A E}

▼

Solve for

A E

CrossMult

Cross multiply

5 (12 - A E) = 13 A E

Distr

Distribute $5$

60 - 5 A E = 13 A E

AddEqn

$LHS + 5 A E = RHS + 5 A E$

60 = 18 A E

DivEqn

$LHS / 18 = RHS / 18$

\frac{6 0}{1 8} = A E

ReduceFrac

$\frac{a}{b} = \frac{a / 6}{b / 6}$

\frac{1 0}{3} = A E

RearrangeEqn

Rearrange equation

A E = \frac{1 0}{3}

CalcQuot

Calculate quotient

A E = 3.33333 \dots

RoundDec

Round to $1$ decimal place(s)

A E \approx 3.3

Now that we know

A E,

we can find

E C

as well.

E C = 12 - A E

Substitute

$A E = 3.3$

E C = 12 - 3.3

SubTerm

Subtract term

E C \approx 8.7

Lengths of $\overline{A F}$ and $\overline{F B}$

Finally, we will find $A F$ and $F B .$

For

A F

and

F B,

we can write the following proportion.

\frac{A C}{B C} = \frac{A F}{F B}

Next, we will again rewrite

F B

in terms of

A F

given that

A B = 5 .

A B = A F + F B ⇓ F B = 5 - A F

Now, let's find

A F .

\frac{A C}{B C} = \frac{A F}{F B}

SubstituteValues

Substitute values

\frac{1 2}{1 3} = \frac{A F}{5 - A F}

▼

Solve for

A F

CrossMult

Cross multiply

12 (5 - A F) = 13 A F

Distr

Distribute $12$

60 - 12 A F = 13 A F

AddEqn

$LHS + 12 A F = RHS + 12 A F$

60 = 25 A F

DivEqn

$LHS / 5 = RHS / 5$

\frac{6 0}{2 5} = A F

ReduceFrac

$\frac{a}{b} = \frac{a / 5}{b / 5}$

\frac{1 2}{5} = A F

RearrangeEqn

Rearrange equation

A F = \frac{1 2}{5}

CalcQuot

Calculate quotient

A F = 2.4

We have found that

A F = 2.4 .

From here, we can also find

F B .

F B = 5 - A F

Substitute

$A F = 2.4$

F B = 5 - 2.4

SubTerm

Subtract term

F B = 2.6

Lengths of BD and DC

Lengths of AE and EC

Lengths of AF and FB

Lengths of $\overline{B D}$ and $\overline{D C}$

Lengths of $\overline{A E}$ and $\overline{E C}$

Lengths of $\overline{A F}$ and $\overline{F B}$

{{ 'ml-heading-exercise' | message }} {{ exercise.name }} {{ 'ml-label-page' | message }} {{ exercise.page }}

{{ 'ml-solutions-missing-heading' | message }}