PROPOSITION XI. THEOREM.

280. The bisector of an exterior angle of a triangle divides the opposite side into external segments having the same ratio as the other two sides.

F

B

Given: In triangle ABC, AD bisecting exterior angle CAF and meeting BC in D;

Through C draw CE to AD, to meet BA in E.

281. COR. Conversely, in ▲ ABC, if D be a point of BC produced, such that

BD: DC, AB : AC,

then AD bisects the exterior / CAF.

For the bisector of that angle must pass through D, and therefore must coincide with AD.

PROPOSITION XII. THEOREM.

282. If three or more transversals passing through the same point make intercepts upon two parallels, these intercepts are proportional.

Given Transversals 04, OB, OC, cutting the parallels AC, A'c', in A, B, C, and A', B', C', respectively;

Through 4', c', draw A'D, C'E, each to OB. Then

As this holds true of the intercepts made by any three transversals through O, it holds true of the intercepts made by any number of such transversals.

283. COR. The parallel intercepts between any two converging transversals are proportional to the intercepts between the parallels and the common point.

For AB: A'B' — AO : A'0 = BO : B'0. (Above)

SIMILAR POLYGONS.

284. Similar polygons are such as are mutually equiangular, and have the sides about the equal angles, taken in the same order, proportional.

285. In similar polygons, similarly situated points, lines, or angles are said to be homologous.

PROPOSITION XIII. THEOREM.

286. Triangles that are mutually equiangular are similar.

Given: In triangles ABC, A'B'C', angle 4 equal to angle 4', angle B to angle B', and angle C to angle c';

To Prove: BC: B'C': = AB: A'B' = AC: A'C'.

Place ▲ A'B'C' upon ▲ ABC, so that A' A,

and AA'B'C' takes the position AB'C'.

287. COR. 1. Two triangles are similar, if two angles of the one are respectively equal to two angles of the other. (121)

288. COR. 2. Two right triangles are similar, if an acute angle of the one is equal to an acute angle of the other. (123)

289. SCHOLIUM. In similar triangles, homologous sides. lie opposite equal angles.

PROPOSITION XIV. THEOREM.

290. Two triangles are similar, if an angle of the one is equal to an angle of the other, and the sides about these angles are proportional.

Given: In triangles ABC, A'B'C', angle 4 equal to angle 4'; also AB: A'B': = AC: A'C';

To Prove Triangle ABC is similar to triangle A'B'C'.

Place A A'B'C' upon ▲ ABC, so that ▲ A' A,

and ▲ A'B'C' takes the position AB'C'.

Then

since AB: A'B' = AC: AC',

(Hyp.)

(276)

(112")

B'C' is to BC,

.. ≤ B = ≤ B', and ≤ C = ≤ c',

.. ▲ ABC is similar to ▲ A'B'C'. Q.E.D. (286)

EXERCISE 357. In the diagram for Prop. XI., if the bisector of angle BAC be drawn so as to meet the base in D', show that BC is divided into internal and external segments having the same ratio.

DEFINITION.-A straight line is said to be divided harmonically if it is divided into internal and external segments having the same ratio; or if it is divided into three segments such that the whole line is to either of its outer segments as the other outer segment is to the inner segment.

358. Show that in the diagram for Exercise 357, BC is divided harmonically according to the first definition above, and BD, according to the second.

359. Show that if D and D' divide any line MN harmonically according to the first definition, then also M and N divide DD' harmonically.

360. Also show that MD is divided harmonically according to the second definition.

PROPOSITION XV. THEOREM.

291. Triangles that have their sides mutually proportional are similar.

Given: In triangles ABC, A'B'C', AB: A'B' = AC : A'C' : BC: B'C';

To Prove Triangle A'B'C' is similar to triangle ABC.

=

On AB, AC, take AD, AE = A'B', A'C', resp., and join DE. Since AB : AD = AC: AE, (Hyp. and Const.)

.. / B' = L D = ≤ B, and ≤ c' = ZE = ZC, (Ax.1)

.. A A'B'C' is similar to ▲ ABC. Q.E.D. (286)

SCHOLIUM. From this and Prop. XI. it is seen that mutually equiangular triangles have their homologous sides proportional; and conversely.