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PROPOSITION III. THEOREM.

313. The bisector of an angle of a triangle divides the opposite side into segments proportional to the other two sides.

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To prove

Let CM bisect the angle C of the trianglè CAB.

MA: MB= CA: CB.

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(if two of a ▲ are equal, the opposite sides are equal).

Putting CA for CE in (1), we have

MA: MB CA: CB.

=

Q. E. D.

PROPOSITION IV. THEOREM.

314. The bisector of an exterior angle of a triangle meets the opposite side produced at a point the distances of which from the extremities of this side are proportional to the other two sides.

E

F

To prove

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Let CM' bisect the exterior angle ACE of the triangle CAB, and meet BA produced at M'.

M'A: M'B = CA: CB.

Proof. Draw AF to CM' to meet BC at F.

Since AF is to CM' of the ▲ BCM', we have

§ 309

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(if two of a ▲ are equal, the opposite sides are equal).

Putting CA for CF in (1), we have

Ax. 1

§ 156

M'A: M'B= CA: CB.

Q. E. D.

315. SCHOLIUM. If a given line AB is divided at M, a point between the extremities A and B, it is said to be divided internally into the segments MA and MB; and if it is divided at M', a point in the prolongation of AB, it is said to be divided externally into the segments M'A and M'B.

M'

M

B

In either case the segments are the distances from the point of division to the extremities of the line. If the line is divided internally, the sum of the segments is equal to the line; and if the line is divided externally, the difference of the segments is equal to the line.

Suppose it is required to divide the given line AB internally and externally in the same ratio; as, for example, the ratio of the two numbers 3 and 5.

х
M'

A

M

B

We divide AB into 5+3, or 8, equal parts, and take 3 parts from A; we then have the point M, such that

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Secondly, we divide AB into two equal parts, and lay off on the prolongation of AB, to the left of A, three of these equal parts; we then have the point M', such that

M'A: M'B = 3:5.

Comparing (1) and (2),

MA: MB = M'A : M'B.

(2)

316. If a given straight line is divided internally and. externally into segments having the same ratio, the line is said to be divided harmonically.

D

317. COR. 1. The bisectors of an interior angle and an exterior angle at one vertex of a triangle divide the opposite side harmonically. For, by §§ 313 and 314, each bisector divides the opposite side into segments proportional to the other two sides of the triangle.

M'

A M

B

318. COR. 2. If the points M and M' divide the line AB harmonically, the points A and B divide the line MM' har

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That is, the ratio of the distances of A from M and M' is equal to the ratio of the distances of B from M and M'.

The four points A, B, M, and M' are called harmonic points, and the two pairs, A, B, and M, M', are called conjugate harmonic points.

SIMILAR POLYGONS.

319. Similar polygons are polygons that have their homologous angles equal, and their homologous sides proportional.

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Thus, if the polygons ABCDE and A'B'C'D'E' are similar

and

the A, B, C, etc., are equal to ▲ A', B', C', etc.

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320. In two similar polygons, the ratio of any two homologous sides is called the ratio of similitude of the polygons.

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In the triangles ABC and A'B'C' let angles A, B, C be equal to angles A', B', C' respectively.

To prove
Proof.

AABC and A'B'C' similar.
Apply the ▲ A'B'C' to the ▲ ABC,

so that A' shall coincide with A.

Then the AA'B'C' will take the position of ▲ AEH.

Now

ZAEH (same as ▲ B') = ▲ B.

.. EH is to BC,

§ 107

(when two straight lines, lying in the same plane, are cut by a third straight line, if the ext.-int. 4 are equal the lines are parallel).

or

.. AB: AE= AC: AH,

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

§ 310

In like manner, by applying A A'B'C' to A ABC, so that LB' shall coincide with Z B, we may prove that

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

323. COR. 2. Two right triangles are similar if an acute angle of the one is equal to an acute angle of the other.

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