Cor. 2. If any number of parallels be drawn cutting two lines, they will divide the lines proportionally.

For, let be the point where AB

If a line divides two sides of a triangle proportionally, it will be parallel to the third side.

Let ABC be a triangle, and let D.E

A

divide AB and AC, so that

AD : ᎠᏴ :: AE : EC;

D

then will DE be parallel to BC.

Draw DC and EB. Then the tri

angles ADE and DEB will have a common altitude; and consequently, we shall have,

ADE : DEB :: AD: DB.

The triangles ADE and EDC have also a common altitude; and consequently, we shall have,

D

A

The antecedents of this proportion being equal, the consequents will be equal; that is, the triangles DEB and

EDC are equal. DE: hence, their the points B and

But these triangles have a common base altitudes are equal (P. VI., C.); that is,

C, of the line BC, are equally distant from DE, or DE prolonged hence, BC and DE are parallel (B. I., P. XXX., C.); which was to be proved.

PROPOSITION XVII. THEOREM.

The line which bisects the vertical angle of a triangle, divides the base into segments proportional to the adjacent sides.

Let AD bisect the vertical angle A of the triangle BAC : then will the segments BD and DC be proportional to the adjacent sides BA and CA.

From C, draw CE parallel to DA, and produce it

are

E

until it meets BA prolonged, at E. Then, because CE and DA are parallel, the angles BAD and AEC are equal (B. I., P. XX., C. 3); the angles DAC and ACE also equal (B. I., P. XX., C. 2). But, BAD and DAC are equal, by hypothesis; consequently, AEC and ACE are equal : hence, the triangle ACE is isosceles, AE being equal to AC.

B

D

In the triangle BEC, the line AD is parallel to the base EC hence (P. XV.),

ВА AE :: BD DC;

or, substituting AC for its equal AE,

ВА : AC :: BD
BD DC;

:

which was to be proved.

PROPOSITION XVIII. THEOREM.

Triangles which are mutually equiangular, are similar.

Let the triangles ABC and DEF have the angle A equal to the angle D, the angle B to the angle E, and the angle C to the angle F: then will they be similar.

point H, of BC; the point D at some point G, of BA;

the side DF will take the position GH, and BGH will

or, since BG is equal to ED, and BH to EF,

hence, the sides about the equal angles, taken in the same order, are proportional; and consequently, the triangles are similar (D. 1); which was to be proved.

Cor. If two triangles have two angles in one, equal to two angles in the other, each to each, they will be similar · (B. I., ·P. XXV., C. 2).

Triangles which have their corresponding sides proportional, are similar.

In the triangles ABC and DEF, let the corresponding aides be proportional; that is, let

AB DE :: BC : EF

CA

FD;

then will the triangles be similar.

For, on BA lay off BG equal to ED; on BC lay

hence, GH is parallel to AC (P. XVI.); and consequently, the triangles BAC and BGH are equiangular, and therefore similar hence,

But, BH is equal to EF; hence, HG is equal to FD. The triangles BHG and EFD have, therefore, their sides equal, each to each, and consequently, they are equal in all their parts. Now, it has just been shown that BHG and BCA are similar hence, EFD and BCA are also similar; which was to be proved.

In

Scholium. In order that polygons may be similar, they must fulfill two conditions: they must be mutually equiangular, and the corresponding sides must be proportional. the case of triangles, either of these conditions involves the other, which is not true of any other species of polygons.