gous; in the second case, the perpendicular sides are homologous.

Cor. 2. The homologous angles are those included by sides respectively parallel or perpendicular to each other.

Scholium. When two triangles have their sides perpenlicular, each to each, they may have a different relative position from that shown in the figure. But we can always construct a triangle within the triangle ABC, whose sides shall be parallel to those of the other triangle, and then the demonstration will be the same as above.

If a straight line be drawn parallel to the base of a triangle, and straight lines be drawn from the vertex of the triangle to points of the base, these lines will divide the base and the parallel proportionally.

Let ABC be a triangle, BC its base, A its vertex, DE parallel to BC, and AF, AG, AII, lines drawn from A to points of the base: then will

DI : BF :: IK : FG :: KL : GH :: LE : HC.

DI : BF :: IK : FG :: KL: GH :: LE : HC;

which was to be proved.

Cor. If BC is divided into equal parts at F, G, and HI, then will DE be divided into equal parts, at I, K,

and L.

PROPOSITION

XXIII. THEOREM.

If, in a right-angled triangle, a perpendicular be drawn from the vertex of the right angle to the hypothenuse : 1o. The triangles on each side of the perpendicular will be similar to the given triangle, and to each other:

2o. Each side about the right angle will be a mean proportional between the hypothenuse and the adjacent segment: 8°. The perpendicular will be a mean proportional between the two segments of the hypothenuse.

1o. Let ABC be a right-angled triangle, A the vertex of the right angle, BC the hypo

thenuse, and AD perpendicular to BC then will ADB and ADC be similar to ABC, and consequently, similar to each other.

The triangles ADB and ABC

B

D

have the angle B common, and the angles ADB and

BAC equal, because both are right angles; they are, there

In like manner, it may be

fore, similar (P. XVIII., C). shown that the triangles ADC and ABC are similar; and since ADB and ADC are both similar to ABC, they are similar to each other; which was to be proved.

2o. AB will be a mean proportional between BC and BD; and AC will be a mean proportional between CB and CD.

For, the triangles ADB and B BAC being similar, their homologous sides are proportional: hence,

BC : AB :: AB : BD.

In like manner,

BC : AC :: AC DC;

which was to be proved.

A

D

3o. AD will be a mean proportional between BL and DC. For, the triangles ADB and ADC being similar,

their homologous sides are proportional; hence,

BD : AD :: AD : DC;

B D

Cor. 2. If from any point A, in a semi-circumference BAC, chords be drawn to to the extremities B and C of the diameter BC, and a perpendicular AD be drawn to the diameter: then will ABC be a right angled triangle, right-angled at A; and from what was proved above, each chord will be a mean proportional between the diameter and the adjacent segment; and, the perpendicular will be a mean proportional between the segments of the diameter.

Triangles which have an angle in each equal, are to each other as the rectangles of the including sides.

Let the triangles GHK and ABC have the angles G and A equal: then will they be to each other as the rectangles of the sides about these angles.

The triangles ADE and ABE have their bases in the same line AB, and a common vertex E; therefore, they have the same altitude, and consequently, are to each other as their bases; that is,

B

ABE : ABC :: AE AC;

multiplying these proportions, term by term, and omitting the common factor ABE (B. II., P. VII.), we have,

ADE : ABC :: AD × AE : AB × AC;

substituting for ADE, its equal, GHK, and for AD x AE, its equal, GH × GK, we have,

GHK : ABC :: GH GK : AB × AC;

which was to be proved.

Cor. If ADE and ABC are similar, the angles D and B being homologous, DE will be parallel to BC, and we shall have, ·

that is, ABE is a mean proportional between ADE and ABC.

B