In the triangles ABC and A'B'C', if LA=LA' and <B=LB', then also LC=LC' and the triangles are similar.

B

The sides AB and A'B' are homologous, so also are the other pairs of sides opposite equal angles.

A

D

с

Let BD through B and B'D' through B' make the
LBDA=LB'D'A'.

Then ABD▲A'B'D' since their angles are respectively equal. In like manner DBC ~ D'B'C', and BD and B'D' divide the triangles similarly.

3. Lines which divide similar triangles similarly are homologous lines of the triangles, and the intersections of homologous lines are homologous points.

Cor. Evidently the perpendiculars upon homologous sides of similar triangles are homologous lines. So also are the medians to homologous sides; so also the bisectors of equal angles in similar triangles; etc.

197°. Theorem.-The homologous sides of similar triangles are proportional.

or

AB BC CA

A'B' B'C' C'A'

=

=

AD. AB AE. AC,

A'C'. AB

A'B'. AC.

B'

B

Proof.-Place A' on A, and let C' fall at D. Then, since LA'=LA, A'B' will lie along AC and B' will fall at some point E. Now, A'B'C'=\\AED, and therefore LAED=2B, and B, D, E, C are concyclic.

Hence

AB AC

A'B'

A'C'

E

=

B

(107°) (176°, 2)

(194°, 2)

Similarly, by placing B' at B, we prove that

AB BC

A'B' B'C'

A

but

Cor. 1. Denoting the sides of ABC by a, b, c, and those of

a b с

A'B'C' by a', b', c',

=

=

a b c

and

and

(195°, 3)

Cor. 2.

b

i.e., the perimeters of similar triangles are proportional to any pair of homologous sides.

B

and

=

AB BC CA

A'B' B'C'

C'A'

a

=

198°. Theorem.--Two triangles which have their sides pro

portional are similar, and have their equal angles opposite homologous sides. (Converse of 197°.)

=

=

Proof.—On A'C' let the ▲A'DC' be constructed so as to have the

LDA'C' = LA
LDC'A' LC.

and

Then

and

a+b+c a+b+c"

AB BC CA

=

A'B' B'C' ̄C'A'

Then LA LA', LB=LB', and
LC=LC'.

▲A'DC'ABC,

AB AC BC

A'D

;
A'C' DC'

AB AC BC
A'B' A'C' B'C''
A'D=A'B'

q.e.d.

LA' LA, LB'=LB,

LC'=LC.

=

=

DC' = B'C',
▲A'DC' =▲A'B'C'.

(196°, Def. 1)

(197°)

(hyp.)

(58°)

q.e.d.

199°. Theorem.-If two triangles have two sides in each proportional and the included angles equal, the triangles are

similar.

AB AC A'B' A'C'

then

and

=

AABCAA'B'C'.

B

Proof.-Place A' on A, and let A'C' lie along AB, and A'B' lie along AC, so that C' falls at D and B' at E.

The triangles AED and A'B'C' are congruent and therefore similar, and

AB AC

AE AD

Hence AB. AD AE. AC; and .. B, D, E, C are concyclic.

and LA=LA',

AB BC A'B' B'C'

AC.

But

1. If BC > AB,

▲AED=2B, and LADE=LC,
ABC AED≈ ▲A'B'C'.

and LA=LA'.

=

200°. Theorem.-If two triangles have two sides in each proportional, and an angle opposite a homologous side in each equal:

1. If the angle is opposite the longer of the two sides the triangles are similar.

2. If the angle is opposite the shorter of the two sides the triangles may or may not be similar.

AABCAA'B'C'.

=

AB BC
AD B'C''

E

B

A

A

E

Proof.—Place A' at A and let B' fall at D, and A'C' along

Draw DE || to BC. Then

▲ABC≈▲ADE, and

...

AB BC

AD DE
DE=B'C'.

(194°) (177°)

(106°, Cor. 3) q.e.d.

=

C A

And since B'C' > A'B', the ▲A'B'C' =▲ADE and they are therefore similar.

(65°, 1)

But

AABCADE,
ΔΑΒC = ΔΑΒ'C'.

2. If BC<AB, B'C' <A'B', and the triangles may or may not be similar.

Proof. Since AD=A'B', and DE B'C', and B'C'<A'B', .. the triangles A'B'C' and ADE may or may not be congruent (65°, 2), and therefore may or may not be similar.

But

AABC AADE,

.. the triangles ABC and A'B'C' may or may not be similar.

Cor. Evidently, if in addition to the conditions of the theorem, the angles C and C' are both less, equal to, or greater than a right angle the triangles are similar.

Also, if the triangles are right-angled they are similar.

201°. The conditions of similarity of triangles may be classified as follows:

1. Three angles respectively equal. (Def. of similarity.) 2. Three sides proportional.

3. Two sides proportional and the included angles equal. 4. Two sides proportional and the angles opposite the longer of the homologous sides in each equal.

If in 4 the equal angles are opposite the shorter sides in each the triangles are not necessarily similar unless some other condition is satisfied.

By comparing this article with 66° we notice that there is a manifest relation between the conditions of congruence and those of similarity.

Thus, if in 3, and 4 of this article the words "proportional" and "homologous " be changed to "equal," the statements become equivalent to 1, 2, and 5 of Art. 66°. The

difference between congruence and similarity is the nonnecessity of equality of areas in the latter case.

When two triangles, or other figures, are similar, they are copies of one another, and the smaller may be brought, by a uniform stretching of all its parts, into congruence with the larger. Thus the primary idea of similarity is that every line-segment of the smaller of two similar figures is stretched to the same relative extent to form the corresponding segments of the larger figure. This means that the tensors of every pair of corresponding line segments, one from each figure, are equal, and hence that any two or more linesegments from one figure are proportional to the corresponding segments from the second figure.

Def.-Two line-segments are divided similarly when, being divided into the same number of parts, any two parts from one of the segments and the corresponding parts from the other taken in the same order are in proportion.

202°. Theorem.-A line parallel to the base of a triangle divides the sides similarly; and

Conversely, a line which divides two sides of a triangle similarly is parallel to the third side.

DE is to AC. Then BA and BC are divided similarly in D and E.

Proof. The triangles ABC and DBE are evidently similar,

AD CE

AB CB
EB'

DB EB'

ᎠᏴ and AB and CB are divided similarly in D and E.

...

=

and ..

A

=

==

E

(195°, 1) q.e.d.

Conversely, if DE so divides BA and BC that AD: DB=CE: EB, DE is || to AC. AD CE AB CB DB EB' DB EB ABC and DBE having the angle B common, and the sides

Proof.—Since

and the triangles