Aim: How do we use ratio, proportion, and similarity effectively?

Do now: If the degree measures of two complementary angles are in the ratio of 2 to 13, what is the degree measure of the smaller angle?

Solution

If 2x = the measure of the smaller angle, then 13x = the measure of the larger angle. Since the two angles are complementary, the sum of their measures is 90.

2x + 13 = 9015x = 9015x/15 = 90/15X = 62x =2(6) = 12And 13x = 13(6) = 78

• A ratio is a comparison by division of two quantities measured in the same units. The ratio of x and y can be written in any one of the following three ways: x/y, x:y, x to y provided x does not equal zero.

• An equation that states that two ratios are equal is called a proportion. To find an unknown member of a proportion, cross-multiply and solve for the term.

Try this:

If ¾ = x/7, find the value of x.

Answer:

Cross multiply:

4x = 21

X = 21/4

Similar Triangles

• When a photograph is enlarged, the original photograph and the enlarged image are similar. Two figures are similar if they have the same shape, but not necessarily the same size. To prove two triangles are similar, it is sufficient to prove that only two pairs of corresponding angles are congruent.

Try this:

The lengths of the sides of a triangle are 9, 15, and 21. If the length of the shortest side of a similar triangle is 12, find the length of its longest side.

Solution

Assume x represents the length of the longest side of the larger triangle. Since the lengths of corresponding sides of similar triangles are in proportion:

Side in smaller triangle/Corresponding side in larger triangle =

9/12 = 21/x

9x = 252

9x/9 = 252/9

X = 28

A

D

B

E

C

Given: DE is parallel to AC.

Prove: Triangle DBE is similar to triangle ABC.

Proof

• DP is parallel to AC.• Angle BDE is

congruent to angle A• Angle BED is

congruent to angle C• Triangle DBE is

similar to triangle ABC

• Given• If 2 lines are parallel,

corresponding angles are congruent

• Same as Reason 2• AA is congruent to

AA

Proving a Proportion

Once two triangles are proved congruent, you may conclude that a pair of corresponding sides are congruent. In much the same way after proving triangles similar, you can write a proportion involving the lengths of corresponding sides as a reason, “Lengths of corresponding sides of similar triangles are in proportion.”

Lets Try It

A

E

CB

D

Given: DE is perpendicular to AC, AB is perpendicular to CD

Prove: EC/BC = ED/AB

Proof

• DE is perpendicular to AC, AC is perpendicular to CD

• Angles 1 and 2 are right angles

• Angle 1 is congruent to angle 2

• Angle C is congruent to Angle C

• Triangle ECD is similar to triangle BCA

• EC/BC = ED/AB

• Given

• Perpendicular lines intersect to form right angles.

• All right angles are congruent.• Reflexive

• AA is congruent to AA

• In similar triangles, corresponding sides are in proportion

Proving Products Equal

To prove the products of two pairs of triangle side lengths are equal, write an equivalent proportion. From the proportion, determine the triangles that must be proved similar. Then prove those triangles similar in the usual way.

A

B L C

DK

Given: ABCD is a parallelogram

Prove: KM x LB = LM x KD

Proof

• ABCD is a parallelogram.

• AD is parallel to BC.

• Angles 1 and 2 are congruent. Angles 3 and 4 are congruent.

• Triangle KMD is similar to triangle LMB.

• KM/LM = KD/LB

• KM x LB = LM x KD

• Given• Opposite sides of a

parallelogram are parallel.• If 2 lines are parallel, then

alternate interior angles are congruent.

• AA is congruent to AA• In similar triangles,

corresponding sides are in proportion

• In a proportion, the product of the means is equal to the product of the extremes.

Proportions in a Right Triangle

If an altitude is drawn to the hypotenuse of a right triangle, the following three theorems are essential.

The two interior triangles formed are similar to each other and to the original triangle.

A B

C

D

III

The altitude is the mean proportional between the measures of the segments in the hypotenuse.

X/m = y/m OR m squared = xy

A

C

BD

y

m

x

The measure of each leg of the original triangle is the mean proportional between the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

H/b = bx or b squared = hx

And

H/a = a/y or a squared = hy

C

A D B

b a

yx

h

In right triangle ABC, altitude CD is drawn to the hypotenuse AB. If AD = 4 and DB = 9, find the lengths of CD, AC, and CB.

A D

C

B94

Solution

Use a Proportion.

Substitute Cross Multiply Find the positive square root of each side.

Left segment/altitude = altitude/right segment

4/CD = CD/9 (CD) squared = 36

CD = 6

Hypotenuse/leg = leg/adjacent segment

13/AC = AC/4 (AC) squared = 4 x 13

AC = 2 (radical 13)

Hypotenuse/leg = leg adjacent segment

13/CB = CB/4 (CB) squared = 9 x 13

CB = 3 (radical 13)