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.

CAH of the ▲ ABC.

=
DB: DC :AB: AC.

To prove

305. COR. If AF be the bisector of the interior angle BAC, we have, from (303) and (304),

BF: FCBD : 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.

EXERCISES.

1. If a b =e: f, and c d e f, show that a : b = c: d.

2. If a b = b: c, show that a: ca2: b2.

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.

SIMILAR FIGURES.

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. (145)

Thus, the figures ABCD, EFGH are similar if

and

ZA ZE, B = ZF, 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',

coincides with A, and A'B' falls on AB.

Let B' fall at

(Hyp.)

(Hyp.)

(78)

(299)

In the same way, by applying the AA'B'C' to the A ABC so that the s at B' and B coincide, we may prove that

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

Combining these two proportions, we have

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

BC: B'C'.

.. the two As are similar.

(307)

Q.E.D.

310. COR. 1. Two triangles are similar when two angles of the one are equal respectively to two angles of the other.

(97)

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',

A A

B

CB'

Since the first ratio in each of these proportions is the same, the second and third ratios in each are equal respectively; that is,

But

... AE = A'C', and DE = B'C'.

... ▲ A'B'C' = AADE,
Δ

having the three sides equal, each to each (108).
Δ ADE is similar to the Δ ABC.
... ▲ A'B'C' is similar to the ▲ ABC.

EXERCISE.

Q.E.D.

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

But

A A

DE is || to BC,

.. the As ABC, ADE are similar.

.. AB : AD = AC: AE,

E

C B'

homologous sides of similar ▲s are proportional (307).

(311)

(Cons.)

(Hyp.)

.. A'C' AE. (Comparing the two proportions)

having two sides and the included equal, each to each (104).

... ▲ A'B'C' is similar to the ▲ ABC.

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.