THEOREM 55 [CONVERSE OF THEOREM 54]. If two triangles have their sides proportional, they are equiangular with one another.

To prove that the As ABC, DEF are equiangular.

Construction Make ▲ FEX = LB and EFX = 4 C, X and D being on

.. the third angles are equal, and the triangles are equiangular.

THEOREM 56. If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.

Hence As DEF, ABC are equiangular, and therefore have their corresponding sides proportional.

.. As DEF, ABC are similar.

Th. 54.

Q. E. D.

NOTE. If DE>AB and DF > AC, H, K lie in AB, AC produced; the proof holds equally well for these cases.

NOTE ON THEOREMS ON SIMILARITY OF TRIANGLES. There is a certain analogy between these three theorems (54, 55, 56) and the three theorems on congruence of triangles (10, 11, 14). Thus:

The phrases "one angle equal" and "three sides proportional" are of course inexact, but serve to show the analogy briefly.

¶Ex. 4. If two triangles have one angle of the one equal to one angle of the other, and the sides about a second angle of each proportional, then the third angles of the triangles are equal or supplementary; and, in the former case, the triangles are similar (compare Note on Ambiguous Case, p. 25).

THEOREM 57. The ratio of the areas of similar triangles is equal to the ratio of the squares on corresponding sides.

Construction

ΔΑΒΕ

BC2

=

AXYZ

YZ2

=

Draw AD to BC, and XW to YZ.

Proof AABC BC. AD, and ▲ XYZ = 1YZ. XW, Th. 26, Cor. 1.

LB=LY (Data), ≤ D= LW (rt. ≤s),

.. the third angles are equal, and the As are equiangular,

COR. 1. If two As ABC, XYZ have only one angle B = one angle Y, then

COR. 2. The ratio of the areas of similar figures is equal to the ratio of the squares on corresponding sides.

NOTE ON THE RATIO OF AREAS OF TRIANGLES.

If two triangles are similar, their areas are proportional to the square of their bases.

If two triangles have the same base, their areas are proportional to their heights.

If two triangles have the same height, their areas are proportional to their bases.

THEOREM 58 (i). The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

AD bisects <BAC internally and cuts BC at D.

Construction Through C draw CE || to DA to cut BA produced at E.

Ex. 5. State and prove the converse of this theorem.

[Use a method of proof analogous to that of Th. 53.]

THEOREM 58 (ii). The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

AD bisects <BAC externally (i.e. AD bisects CAF) and cuts BC produced at D.

Construction Through C draw CE || to DA to cut BA at E.