CASE. II. — When A E and E B (Fig. 2) are incommensurable.

Divide A E into any number of equal parts, and apply one of these parts to E B as often as it will be contained in E B.

Since A E and E B are incommensurable, a certain number of these parts will extend from E to a point K, leaving a remainder K B, less than one of the parts.

Draw K H Il to BC.
Since A E and E K are commensurable,
EK FH .

(Case I.) A E – AF Suppose the number of parts into which A E is divided to be continually increased, the length of each part will become less and less, and the point K will approach nearer and nearer to B. The limit of E K will be E B, and the limit of FH will be FC.

.. the limit of E K will be E B.

AE'

and

the limit of %

will be

A

.: their limits EB

and Fa

EK.no FH
Now the variables and are always equal, how-

A E
ever near they approach their limits ;
. .. their limits

are equal, § 199

Q. E. D. 276. COROLLARY. One side of a triangle is to either part cut off by a straight line parallel to the base, as the other side is to the corresponding part. Now EB : A E ::FC : A F.

§ 275 By composition,

EB + A E : AE :: FC + AF: AF, § 263 or,

A B : AE :: AC : AF.

PROPOSITION III. THEOREM.

277. If a straight line divide two sides of a triangle proportionally, it is parallel to the third side.

BC

In the triangle A B C let EF be drawn so that

AB AC

A E AF
We are to prove EF || to BC.

From E draw E H || to BC.
Then

A B AC
A E - AH'

$ 276

(one side of a A is to either part cut off by a line Il to the base, as the other

side is to the corresponding part).

A B A C
But
A E – AF

Hyp. · AC AC

Ax. 1 "AF AH'

.. AF = AH.

.. E F and E H coincide,

(their extremitics being the same points).
But
E H is ll to BC;

Cons. i. E F, which coincides with E H, is II to BC.

Q. E. D. 278. Def. Similar Polygons are polygons which have their homologous angles equal and their homologous sides proportional.

Homologous points, lines, and angles, in similar polygons, are points, lines, and angles similarly situated.

ON SIMILAR POLYGONS.

PROPOSITION IV. THEOREM. 279. I'wo triangles which are mutually equiangular are similar.

BIL

B4
In the A ABC and A' B'C' let A, B, C be equal to

S A', B', C' respectively.
We are to prove A B : A' B' = AC : A'C' = BC : B'C'.

Apply the A A'B'C' to the A A BC,

so that 2 A' shall coincide with 2 A.
Then the A A' B' C' will take the position of A A E H.
Now ZA EH (same as 2 B') = 2 B.
... E H is ll to BC,

$ 69 (when two straight lines, lying in the same plane, are cut by a third straight line, if the ext. int. & be equal the lines are parallel). . AB : AE = AC : A H,

$ 276 (one side of a A is to either part cut off by a line Il to the base, as the other

side is to the corresponding part).
Substitute for A E and A H their equals A' B' and A'C'.
Then A B : A' Bi = AC : A'C'.
In like manner we may prove

A B : A' B' = BC : B'C'.
.. the two A are similar.

$ 278

Q. E. D. 280. Cor. 1. Two triangles are similar when two angles of the one are equal respectively to two angles of the other.

281. Cor. 2. Two right triangles are similar when an acute angle of the one is equal to an acute angle of the other.

PROPOSITION V. THEOREM. 282. Two triangles are similar when their homologous sides are proportional.

AL

A

In the triangles A B C and A' B' C' let

AB AC BC

A' B A C - Ᏼ Ꮯ "
We are to prove
& A, B, and C equal respectively to E A', B', and C'.

Take on A B, A E equal to A' B',
and on A C, A H equal to A'C'. Draw E H.

A B AC
A'B' — A'C'

Hyp. Substitute in this equality, for A' B' and A' C their equals A E and A H.

A B AC
Then

A È – AH
.. E H is ll to BC,

§ 277 (if a line divide two sides of a A proportionally, it is ll to the third side). Now in the A A BC and A EH ZA BC= LA EH,

§ 70 (being ext. int. angles). LACB= L A HE,

$ 70 ZA= 2 A.

Iden. .. A A B C and A E H are similar, $ 279 (two mutually equiangular S are similar). . AB AE

§ 278 ''BC - EH (homologous sides of similar A are proportional).

$ 108

But

A B A' B'
BC – B'C"

§ 278
. A E A' B
EH = B'C'

Ax. 1
Since
A E = A' B',

Cons.
EH= B'C'.
Now in the A A E H and A' B'C',
EH= B'C', A E= A' B', and A H = A' C',

. A A E H= A A' B'C', (having three sides of the one equal respectively to three sides of the other). But A A E H is similar to A ABC. i. A A' B'C' is similar to A A BC.

Q. E. D. . 283. ScHolium. The primary idea of similarity is likeness of form; and the two conditions necessary to similarity are :

I. For every angle in one of the figures there must be an equal angle in the other, and

II. the homologous sides must be in proportion.

In the case of triangles either condition involves the other, but in the case of other polygons, it does not follow that if one condition exist the other does also.

Thus in the quadrilaterals Q and Q', the homologous sides are proportional, but the homologous angles are not equal and the figures are not similar.

In the quadrilaterals R and R', the homologous angles are equal, but the sides are not proportional, and the figures are not similar.