SOLUTION.-There are several ways of solving this problem in practice. The one illustrated in the figure is as follows: Any convenient distance AC is measured and the angle C observed with a transit or compass. From C, a distance CĘ is measured, and at E an angle AED equal to C is turned off, The point D where the line of sight ED meets A B is marked, and the distances AD and AE are measured. Then, since AED equals C, the lines ED and CB are parallel (see Geometry, Part 1) and, therefore (Art. 9),

AB: AC = AD: AE

B

EXAMPLE 2.-Divide a line AB, Fig. 1, of given length into two parts A E and E B whose ratio shall be the same as that of two given numbers m and n; that is, so that AE : E B = m : n.

SOLUTION.-Since A E: EB = m:n, we must have (Art. 6),

AE can be found in a similar manner, or by subtracting the value of E B from AB.

EXAMPLES FOR PRACTICE

=

100 feet, A E

1. If the measured distances in Fig. 4 are A C = 45.2 feet, AD = 48.36 feet, what are the distances A B and D B?

3. If AB, Fig. 1, is equal to 125 feet, find the distances A E and

EB so that the line will be divided at E in the ratio of 5 to 2.

12. If two lines, as A B and CD, Fig. 5, are cut by any number of parallel lines, as EM, GN, IO, etc., the corresponding intercepts are proportional; that is, EG: GI = MN: NO; GI: IK NO: OP, or, by interchanging the means, EG: MN GI: NO = IK: OP, etc.

=

A

K

J

L

B

FIG. 5

or,

=

=

=

=

MP MN ΝΟ ОР

EK: MP = EG: MN = GI: NO = IK: OP

13. In any triangle A B C, Fig. 6, the bisector BD of an angle divides the side opposite proportionally to the including sides; that is, A B: BC= AD: DC.

Draw CE parallel to B D and meeting A B produced in E. Then, in the triangle A E C, by Art. 9,

Substituting, in equation (1), B C for its equal B E,

A

POLYGONS

SIMILAR POLYGONS

SIMILAR TRIANGLES

14. Similar polygons are those whose corresponding angles are equal and whose corresponding sides are proportional.

In order that two polygons may be similar, it is manifestly necessary that each angle of the one shall be equal to the corresponding angle of the other. But this is not sufficient; the corresponding sides must be proportional. For example,

the quadrilaterals ABCD and A'B'C' D', Fig. 7, have their corresponding angles equal, but they are not similar, because their corresponding sides are not proportional. The quadrilaterals ABCD and A" B" C" D" have their corresponding angles equal and their corresponding sides proportional, and are, therefore, similar.

The corresponding sides of similar polygons are called homologous sides.

15. Two triangles are similar when the angles of one are equal to the angles of the other.

B

In Fig. 8, let the angles of the triangle A B C be equal, respectively, to those of the triangle A'B' C'. Place the triangle A' B' C' upon ABC, so that the angle A' will coincide with its equal A. Then B' will fall along AB and C'along AC, as at B" and C", respectively, and B' C' will take the position B" C". The angle B", which is equal to B', is equal to B, and the angle C", which is equal to C', is equal to C; hence, B" C" is parallel Then, by Art. 10,

B"

C"

B'

C

FIG. 8

to BC (see Geometry, Part 1).

Therefore, the triangles, having their angles equal and their corresponding sides proportional, are similar.

16. Two triangles are similar when two angles of the one are equal respectively to two angles of the other.

17. Two right triangles are similar when an acute angle of one is equal to an acute angle of the other.

18. A triangle is similar to any triangle formed by a line parallel to one of its sides and the segments it intercepts on the other two sides of the other two sides prolonged.

19. Two triangles are similar when the three sides of one are either parallel or perpendicular to the three sides of the other.

20. Two triangles are similar when their corresponding sides are proportional.

In Fig. 9,

AB: A'B' = AC: A'C' = BC: B' C

But,

AB: AD = BC: DE, or AB : A' B' = BC: DE

The last two proportions are the same, term for term, excepting the last term; hence, DE is equal to B' C', and the triangles A DE and A' B' C' are equal. Therefore, the triangles A B C and A' B' C' are similar.

21. Two triangles are similar when an angle of the one is equal to an angle of the other and the including sides are proportional.

22. In two similar triangles, corresponding altitudes have the same ratio as any two corresponding sides. Let CD and C'D', Fig. 10, be the corresponding altitudes of the

BC: B'C'

CD:CD = AC: A' C' = AB: A'B'

23. As stated in Art. 14, two polygons are similar when their corresponding angles are equal and their corresponding sides are proportional. It has now been shown that, in triangles, either of these conditions includes the other. This could have been expected from the fact that either the three