Exercise 33 Page 432

Let's consider the following diagram.

We want to determine EF. To do so we will use the fact that all pairs of corresponding sides of two similar figures are proportional. First let's find AE and EC.

We can see that ∠ DEA and ∠ CEB are vertical angles, so ∠ DEA≅∠ CEB. Since AD∥BC, ∠ ADE≅∠ EBC by the Vertical Angles Congruence Theorem. Two angles of △ AED are congruent to two angles of △ CEB, so △ AED and △ CEB are similar by the Angle-Angle (AA) Similarity Theorem. △ AED~△ CEB Note that AD corresponds to BC, and AE corresponds to EC. Recall that all pairs of corresponding sides of two similar figures are proportional. AD/BC=AE/EC We know that AD= 40 and BC= 30. Let's substitute these values into the above equation and determine AE in terms of EC.

AD/BC=AE/EC

SubstituteII

AD= 40, BC= 30

40/30=AE/EC

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Solve for AE

RearrangeEqn

Rearrange equation

AE/EC=40/30

ReduceFrac

a/b=.a /10./.b /10.

AE/EC=4/3

MultEqn

LHS * EC=RHS* EC

AE=4/3EC

We have that AE is equal to four-thirds of EC. Let's also find AC in terms of EC.

AC=AE+EC

Substitute

AE= 4/3EC

AC= 4/3EC+EC

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Add terms

Rewrite

Rewrite EC as 1* EC

AC=4/3EC+1* EC

OneToFrac

Rewrite 1 as 3/3

AC=4/3EC+3/3EC

AddFrac

Add fractions

AC=7/3EC

We have that AC is equal to seven-thirds of EC. Now we will move on to determining EF. First we have to find a triangle similar to △ FCE.

Angles ∠ EFC and ∠ ADC are both right angles, so they are congruent. Triangles △ FCE and △ DCA share an angle, ∠ FCE. In conclusion, we have that two angles of △ FCE are congruent to two angles of △ DCA, so △ FCE is similar to △ DCA by the AA Similarity Theorem. △ DCA~△ FCE Note that EF corresponds to AD, and EC corresponds to AC. Remember that AD= 40 and AC=73EC. EF/AD=EC/AC ⇔ EF/40=EC/.7 /3.EC Let's solve the obtained equation for EF.

EF/40=EC/.7 /3.EC

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Solve for EF

ReduceFrac

a/b=.a /EC./.b /EC.

EF/40=1/.7 /3.

DivByFracD

a/b/c= a * c/b

EF/40=3/7

MultEqn

LHS * 40=RHS* 40

EF=120/7