bisectors of the sides AB, AC, and BC, respectively, of ABC.

Conclusion. DM, EN, and FP are concurrent at a point equidistant from A, B, and C.

Suggestions. First prove that DM and EN intersect at a point O by § 55. Then show that O is equidistant from A and B and from A and C, and hence from Band C. Why does it follow that O is on FP?

NOTE. - The point of intersection of the perpendicular bisectors of the sides of a triangle is called the circumcenter of the triangle.

113. Theorem. - The three altitudes of any triangle are

Hypothesis. AD, BE, and CF are the altitudes of △ABC. Conclusion. AD, BE, and CF are concurrent.

Suggestions. Draw MN || AB through C, MP || BC through A, and PN AC through B, forming △ MNP. Prove that AD, BE, and CF are perpendicular bisectors of the sides of AMNP. Το prove that is the middle point of MN, show that MC=AB = CN. How may it be proved that A is the middle point of MP and B the middle point of PN? Why are the altitudes of △ABC perpendicular to the sides of MNP? Why then are the altitudes of ∆АВС concurrent?

Write the proof in full.

NOTE. - The point of intersection of the altitudes of a triangle is called the orthocenter of the triangle.

114. Theorem. — The medians of any triangle are concurrent at a point of trisection of each.

Hypothesis. AD, BE, and CF are the medians of ∆ABC. Conclusion. AD, BE, and CF are concurrent at a point of trisection of each.

Suggestions. Show by § 56 that AD and BE meet at a point O. Let G and H be the middle points of AO and BO, respectively. Draw GH, HD, DE, and EG. If GHDE is proved a parallelogram, EO = OH = HB and AG=G0 = OD. Why? Hence, AD and BE meet at a point of trisection of each. Similarly, it may be proved that CF and BE meet at a point of trisection of each, and hence CF passes through 0.

Write the proof in full.

NOTE. - The point of intersection of the medians of a triangle is called the centroid of the triangle.

EXERCISES

1. Prove that the perpendicular bisectors of the legs of a right triangle intersect on the hypotenuse.

2. Where do the three perpendicular bisectors of the sides of an obtuse triangle intersect? Draw a figure to illustrate.

3. Prove that the altitudes of a right triangle intersect at the vertex of the right angle.

4. Where do the altitudes of an obtuse triangle intersect? Draw a igure to illustrate.

5. Prove that the perpendicular bisectors of the four sides of a rectangle are concurrent.

6. Prove that the perpendicular bisectors of the four sides of an isosceles trapezoid are concurrent.

7. Construct a circle which shall pass through the vertices of a given triangle.

8. Prove that the bisectors of the exterior angles at two vertices and the bisector of the interior angle at the third vertex of any triangle are concurrent.

A

C

B

0

E

FD

SUGGESTION. - The proof is similar to that in § 111. NOTE. - The point O of intersection is called an excenter of the triangle. Every triangle has three excenters.

9. Prove the theorem in § 114 by the following construction: Show that medians AD and BE intersect at a point O. Then draw CF through O, meeting AB at F, and produce it to G, so that OG = CO. Draw AG and BG. Show that AGBO is a parallelogram, and hence that A AF = FB.

C

E

D

0

F

B

G

10. If D is the orthocenter of △ABC, prove that A is the orthocenter of A BCD, B is the orthocenter of ACD, and C is the orthocenter of AABD.

CHAPTER VI

PROPORTION: SIMILAR POLYGONS

115. Numerical measures of geometric figures. - The numerical measure of a geometric figure is the number of times that it contains a given unit of measure.

Thus, if the line-segment MN contains the unit of measure x six times, M

its numerical measure is 6.

N

Similarly, if ∠ABC contains exactly 30°, its numerical measure, when the unit of measure is a degree, is 30.

116. Ratio of geometric figures. - The ratio of two geometric figures of the same kind is the quotient of their numerical measures, when the same unit of measure is applied to each. Thus, if the line-segment AB contains a unit of measure 3 times, and the line-segment CD contains it 5 times, the ratio of AB to CD A is 3 divided by 5, or .

C

B

60

D

If one of two angles contains 40° and the other one 60°, the ratio of the first to the second is. In any ratio the dividend is called the antecedent and the divisor is called the consequent.

117. Proportion. — An expression of equality between two ratios is called a proportion.

a

The proportion =, being an equation, is read either as

an equation "a is to base is to d."

or

In any proportion

a

=

C

the numbers a, b, c, and d are called the first, second, third, and fourth terms, respectively. The first and fourth terms, a and d, are called the extremes. The second and third terms, b and c, are called the means.

The four terms of a proportion are said to be "proportional," or "in proportion."

The fourth term of a proportion is called the fourth proportional to the other three taken in order. Thus, in the proportion=, d is the fourth proportional to a, b, and c.

A proportion in which the means are equal is called a mean proportion, and either mean term is called the mean proportional between the extremes. Thus, in the proportion

m = x,

n

, x is a mean proportional between m and n.

It must be anderstood from the definition of a proportion that the terms are abstract numbers. Hence, since a proportion is an equation, all processes which may be performed upon an equation may be performed upon a proportion.

20. 12 x2 and 3 y2.

22. 4 and (a – b)2.

23. x2 and (x + 1)2.