Cor. 2. A triangle having two equal angles, is isosceles (106) (105).

Cor. 3. An equiangular triangle is equilateral (106).

(107)

What is an Isosceles triangle? What an Equilateral triangle?

PROPOSITION III.

Any two sides of a triangle are together greater than (108) the third.

Let ABC be any triangle; then will the sum of any two sides, as AC+ CB, be > the third AB. Produce AC to D, making CD = CB, and join

.. (105) AB< AD = AC+ CD = AC + CB. Q. E. D.

Fig. 16.

Cor. 1. The sum of the lines joining any point within a (109) triangle and the extremities of one of its sides, is less than the sum of the other two sides.

Let D be any point in the triangle ABC; produce AD to meet CB in E, then

AD+DBAD + DE + EB < AC+CE + EB. (Why?) Q. E. D.

E

B

Fig. 162.

Cor. 2. Any side of a polygon is less than the sum of all (110) the other sides.

[Letter and show how.]

Fig. 163.

.. Cor. 3. Of two polygons, standing upon the same (111) base and one embracing the other, the perimeter of the enveloping is greater than that of the enveloped.

[See the last figure and explain.]

Cor. 4. The straight line is the shortest that can be (112) drawn from one point to another.

P

First. Let APB be any line, whether curved or broken, but concave toward the straight line AB, then APB will be greater than AB. For, AR taking any point P, in the line APB, and revolv

E

Fig. 164.

SHORTEST DISTANCE.

75

ing the branches AP, BP, about A and B till the point P coincides with the straight line AB, first in Q, then in R, the straight lines AP, BP, in the positions AQ, BR, will overlap each other by a certain line RQ, since the st. line AP + st. 1. BP > AB—and therefore, the branches AP, BP, falling on the same side of AB, since, by hypothesis, the path APB is concave toward AB, will intersect each other in some point E; whence the path APB, being equal to the sum of the paths AE+ EQ and RE+ EB, will be longer than the path AEB by the sum of the parts RE and QE. The same may be shown of any other path that is concave toward AB except AB.

Second. Let the path AcdefghB be any what

ever, having flexures either continuous or abrupt
in c, d, e, f, g, h; then drawing the straight lines A

Ac, cd, de, ef, fg, gh, hB,

Fig. 165. we have (110) AB <polygonal perimeter (Ac+cd+de+ ef +fg+gh+hB) < the path (AcdefghB),

as shown above, since all the parts of AcdefghB are concave toward the several sides of the polygon. Q. E. D.

Cor. 5. "Hence also, we may infer, that of any two (113)

C

D

paths, ACB, ADB, leading from A to B, and everywhere concave toward the straight line AB, that which is enveloped by the other, as ADB, is the A shorter. For of all the paths not lying between Fig. 166. ADB and the straight line AB, there is none, ADB excepted, than which a shorter may not be found. And this is the case whether the paths ACB, ADB, be both of them curvilineal, or one of them, (ACB or ADB), rectilineal.” *

Cor. 6. The perpendicular is the shortest distance from (114) a point to a straight line; and of oblique lines, that is the shorter which is nearer the perpendicular-and those equally distant are equal.

For let PA be perpendicular to CBAB', and AB' = AB AC, and produce PA in Q, making AQ = AP; joining QB, QB', QC, we have

PC+QC> PB+QB > PA+QA (?)

BA

B

Cor. 7. All points equally distant from the extremities (115) of a straight line are situated in the same perpendicular passing through the middle of that line.

The points C, B, A, B', are equally distant from P and Q.

PROPOSITION IV.

If two triangles have two sides of the one equal to two (116) sides of the other, each to each, but the included angles unequal, then the third sides will be unequal, and the greater side will be that opposite the greater angle.

Let two of the equal sides be made common in AB, and the other two, AC, AD lie on opposite sides of AB; draw AI bisecting the angle CAD and terminating in I, which will be a point in CB on the hypothesis that CAB is the greatFig. 17. er angle, since then we have CAB > CAI = DAI > DAB ;— finally, join ID. We have (96), (108),

BC= BI+IC = BI + ID > BD. Q. E. D.

Cor. 1. Conversely, if two triangles have two sides of (117) the one equal to two sides of the other, each to each, but the third sides unequal, the opposite angles will be unequal and that will be the greater which is opposite the greater side. For it can be neither equal nor less. (Why?)

Cor. 2. If two triangles have the sides of the one seve- (118) rally equal to the sides of the other, the angles opposite the equal sides will be severally equal.

For, if AB = A'B', BC = B'C', AC = A'C' ;
../C cannot be C', for then would ABA'B';
.. < C = C', and, for a like reason, A = A', A
B = B'.

Β Α'
Fig. 172.

Cor. 3. When the opposite sides of a quadrilateral are (119) equal, each to each, the figure will be a parallelogram [fig. 14,].

P'

1o. It is a principle in Optics that a ray of light, AP, impinging upon a reflecting surface, MM', makes the angle of reflection BPP', equal M to the angle of incidence APP', PP' being perpendicular to MM' at P. Given in position the luminous point A, the eye, B, and the mirror, MM', to find the point P from which the light is reflected to B, and to prove that APB is the shortest path from A to B by way of the mirror, or that AP+PB < AQ + QB.

2o. Having given in position the elastic planes PQ, QR, and the points A, B; it is required to find the track of an elastic ball projected from A upon PQ and rebounding from QR so as to hit a pin at B.

P

3°. Given in position the straight line MN and the points A, B; required the point O in MN such that AO shall BO.

Fig. 18.

Α

Fig. 182

B

R

Fig. 19.

.

4°. Prove that the perpendiculars drawn through the middle points of the three sides of a triangle, will intersect in the same point.

5°. Having given in position the lines AB, CD, and the point P, it is required to draw through P a line mn, which shall make equal angles with AB, CD.

6°. Let AOB be a right angle, COB an equilateral triangle, and OP perpendicular to CB. Prove that the angles AOC, COP, POB are all equal. How could you trisect a right angle?

7. Let A, B, two angles of an equilateral triangle, be bisected by the lines AO, BO, and from O draw OP, OQ, parallel to the sides CA, CB, and terminating in the base in P and Q. Prove that AP = PQQB. [The student will construct the figure.]

8°. Prove that any side of a triangle is greater than the difference of the other two.

9°. Prove that the angle included between two lines drawn from the vertex of any triangle, the one bisecting the vertical angle and the other perpendicular to the base, is half the difference of the basic angles.

10°. Prove that if a line be drawn from one of the equal angles of an isosceles triangle upon the opposite side and equal to it, the angle embraced by this line and the production of the base, or unequal side, will be three times one of the equal angles of the triangle.

11. Draw three lines from the acute angles of a right angled triangle-two bisecting these angles and the third a perpendicular to one of the bisecting lines-and prove that the triangle embraced by these lines will be isosceles.

SECTION THIRD.

Proportional Lines.

PROPOSITION I.

The segments of lines intercepted by parallels are pro- (120) portional.

C

Fig. 23.

B'

In the first place, let the secant line AC be divided into commensurable parts by the parallels AA', BB', CC', for instance, such that AB containing three measures, BC shall contain two of the same; it be may shown that any other secant line, A'B'C' will be divided in the same ratio. For, dividing AB, BC, into 3 and 2 parts in a, b, c, and drawing aa', bb', cc', through the points of division parallel to AA' and through A', a', b', &c. (points of A'C'), drawing A'a", a'b", &c., parallel to AC and terminating severally in the parallels aa', bb', &c., we have (?) (?),