because each is a right angle; they are, therefore, similar (P. XVIII., C.). In like manner, it may be shown that the triangles ADC and ABC are similar; and since ADB and ADC are each similar to ABC, they are similar to each other; which was to be proved.

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

B

For, the triangles ADB and 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.

3°. AD is a mean proportional between BD and DC. For, the triangles ADB and ADC being similar, their homologous sides are proportional; hence,

Cor. 2. If from any point A, in a semi-circumference
BAC, chords are drawn to the extremi-

ties B and C of the diameter BC, and
a perpendicular AD is drawn to the
diameter: then ABC is a right-angled
triangle, right-angled at A; and from
what was proved above, each chord

a

is

mean proportional between the diameter and the adjacent segment; and, the perpendicular is a

tional between the segments of the diameter.

mean propor

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 are they to
each other as the rectangles
of the sides about these
angles.

For, lay off AD equal to
GH, AE to GK, and draw DE;
then the triangles ADE and
GHK are equal in all respects.

Draw EB.

K

E

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,

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

ADE : ABC :: AD X AE : ABX AC;

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

GHK : ABC :: GH X GK : ABX AC,

which was to be proved.

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

Similar triangles are to each other as the squares of their homologous sides.

Let the triangles ABC and DEF be similar, the angle A being equal to the angle D, B to E, and C to F: then the triangles are to each other as the squares of any two homologous sides.

Because the angles A and D are equal, we have (P.

combining this with the first proportion (B. II., P. IV.),

In like manner, it may be shown that the triangles are to each other as the squares of AB and DE, or of BC and EF; which was to be proved.

Similar polygons may be divided into the same number of triangles, similar, each to each, and similarly placed.

Let ABCDE and FGHIK be two similar polygons, the angle A being equal to the angle F, B to G, C to H, and so on: then can they be divided into the same number of similar triangles, similarly placed.

Because the polygons are similar, the triangles ABC and FGH have the angles B and G equal, and the sides about these angles proportional; they are, therefore, similar (P. XX.). Since these triangles are similar, we have the angle ACB equal to FHG, and the sides AC and FH, proportional to BC and GH, or to CD and HI. The angle BCD being equal to the angle GHI, if we take from the first the angle ACB, and from the second the equal angle FHG, we have the angle ACD equal to the angle FHI: hence, the triangles ACD and FHI have an angle in each equal, and the including sides proportional; they are therefore similar.

In like manner, it may be shown that ADE and FIK are similar; which was to be proved.

Cor. 1. The corresponding triangles in the two polygons are homologous triangles, and the corresponding diagonals are homologous diagonals.