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Proposition 15. Theorem.
304. The bisector of an exterior angle of a triangle divides the opposite side externally into segments proportional to the adjacent sides. Hyp. Let AD bisect the ext.
H Z CAH of the A ABC.
To prove DB : DC = AB: AC.
Proof. Draw CE | to DA meeting BA at E. Since DA is || to CE, (Cons.)
305. CoR. If AF be the bisector of the interior angle BAC, we have, from (303) and (304),
BF: FC = BD : CD.
306. DEF. When a straight line is divided internally and externally into segments having the same ratio, it is said to be divided harmonically. Therefore (305),
The bisectors of an interior and exterior angle at the vertex of a triangle divide the opposite side harmonically.
1. If a :b = e:f, and c:d =e: f, show that a : b=
2. If a
6 b:c, show that a:c= a?: . 3. A, B, C, D are four points on a straight line, E is any point outside the line. If the parallelograms AEBF, CEDG be completed, show that FG will be parallel to AD.
307. DEF. Similar figures are those which are mutually equiangular (144), and have their homologous sides proportional.
The homologous sides are those which are adjacent to the equal angles.
Thus, the figures ABCD, EFGH are similar if
LA = _E, ZB = _F, ZC = _G, etc.,
*The sides AB and EF are homologous, since they are adjacent to the equal angles A and B, and E and F, respectively; also the sides BC and FG are homologous, etc. The diagonals AC, EG are homologous.
308. The constant ratio of any two homologous sides of similar figures is called the ratio of similitude of the figures.
Proposition 16. Theorem. 309. Triangles which are mutually equiangular are similar.
Hyp. In the As ABC, A'B'C', let LA= L A', ZB= LB', 2C= _0'.
A ABC similar to
B to the A ABC so that the pt. A' coincides with A, and A'B' falls on AB. Let B'fall at D. Then, since ZA' = LA,
(Hyp.) ... A'C' falls on AC. Let
C' fall at E; then B'C' falls on DE.
(Hyp.) .:. DE is || to BC.
(78) .:. AB : AD = AC: AE,
(299) AB : A'B' = AC: A'C'. In the same way, by applying the AA'B'C' to the A ABC so that the Zs at B' and B coincide, we may
AB : A'B' = BC: B'C'.
AB : A'B' = AC: A'C' = BC : B'C'.
Q.E.D. 310. Cor. 1. Two triangles are similar when tivo angles of the one are equal respectively to two angles of the other.
311. COR. 2. A triangle is similar to any triangle cut off by a line parallel to one of its sides.
312. Sch. In similar triangles the homologous sides lię opposite the equal angles,
Proposition 17. Theorem. 313. Triangles which have their homologous sides proportional are similar. Hyp. In the As ABC, A'B'C',
АВ AC BC let A'B' A'C'
... the As ABC, ADE are similar. (311)
AB AC BC
(Cons.) AB AC BO AD A'C B'C'.
(Hyp.) Since the first ratio in each of these proportions is the same, the second and third ratios in each are equal respectively; that is, AC AC
DE B'C' -
.:. AE = A'C', and DE =
.:: A A'B'C' = AADE, ,
EXERCISE. If the sides of the triangle in Prop. 14 are AB = 8, BC = 12, and AC = 10, find the lengths of the segments BD and CD.
Proposition 18. Theorem. 314. Two triangles which have an angle of the one equal to an angle of the other, and the sides about these angles proportional, are similar. Hyp. In the As ABC, A'B'C', let
DE is 1 to BC,
:. AB : AD = AC: AE, homologous sides of similar As are proportional (307). But A'B' = AD.
(Cons.) .:. AB: AD = AC: A'C'.
(Hyp.) . A'C' = AE. (Comparing the two proportions)
.:. A A'B'C' = A ADE,
Q.E.D. 315. Sch. From the definition (307), it is seen that two conditions are necessary that polygons may be similar: (1) they must be mutually equiangular, and (2) their homologous sides must be proportional. In the case of triangles we learn from Props. 16 and 17 that each of these conditions follows from the other; so that one condition is sufficient to establish the similarity of triangles.
This, however, is not necessarily the case with polygons of more than three sides; for even with quadrilaterals, the angles can be changed without altering the sides, or the proportionality of the sides can be changed without altering the angles.