Exercise 35 Page 579

Let's analyze the given figure. Since we are given a triangle with a line that is parallel to one of its sides, we can use the Triangle Proportionality Theorem.

The lengths of the segments intercepted by the parallel line are proportional. Let's write a proportion using the expressions for the lengths of the segments. CD/DA=CE/EB ⇕ 2/DA=t-2/t+1Since we are given the lengths of CD and CA, we can find the length of DA.

CD+DA=CA

SubstituteII

CD= 2, CA= 10

2+DA= 10

SubEqn

LHS-2=RHS-2

DA=8

Now we can substitute the missing length into our proportion. 2/DA=t-2/t+1 ⇕ 2/8=t-2/t+1 Let's solve it for t.

2/8=t-2/t+1

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Solve for t

MultEqn

LHS * (t+1)=RHS* (t+1)

2/8(t+1)=t-2

ReduceFrac

a/b=.a /2./.b /2.

1/4(t+1)=t-2

MultEqn

LHS * 4=RHS* 4

t+1=4(t-2)

Distr

Distribute 4

t+1=4t-8

AddEqn

LHS+8=RHS+8

t+9=4t

SubEqn

LHS-t=RHS-t

9=3t

DivEqn

.LHS /3.=.RHS /3.

3=t

RearrangeEqn

Rearrange equation

t=3

Finally, we will find CE.

CE=t-2

Substitute

t= 3

CE= 3-2

SubTerm

Subtract term

CE=1

We found that t=3 and CE=1.