1. 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?

2. 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 = 90 15x = 90 15x/15 = 90/15 X = 6 2x =2(6) = 12 And 13x = 13(6) = 78

3. 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.

4. Try this: If ¾ = x/7, find the value of x.

5. Answer: Cross multiply: 4x = 21 X = 21/4

6. 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.

7. 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.

8. 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

9. A D B E C Given: DE is parallel to AC. Prove: Triangle DBE is similar to triangle ABC.

10. 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

11. 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.”

12. Lets Try It A E C B D Given: DE is perpendicular to AC, AB is perpendicular to CD Prove: EC/BC = ED/AB

13. 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

14. 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.

15. A B LC D K Given: ABCD is a parallelogram Prove: KM x LB = LM x KD

16. 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.

17. Proportions in a Right Triangle If an altitude is drawn to the hypotenuse of a right triangle, the following three theorems are essential.

18. The two interior triangles formed are similar to each other and to the original triangle. A B C D III

19. 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 B D y m x

20. 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 ba yx h

21. 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 B 94

22. Solution Use a Proportion. SubstituteCross MultiplyFind 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)