Fig. 268.

A B C D: B E D F:: Cd: Df.

The areas in this particular case are as 4 to 3. But in general the proportion will be as the lengths of the bases. Thus the proposition is proved in regard to rectangles, but it has been shown (Art. 341) that all parallelograms of equal base and altitude are equal. Therefore the proposition is proved in regard to parallelograms generally, including rectangles.

## 360. - Proposition

Triangles of equal altitude are to each other as their bases. It has been shown (Art. 359) that parallelograms of equal altitude are in proportion as their bases, and it has also been shown (Art. 342) that of a triangle and parallelogram, when of equal base and altitude, the parallelogram is equal to double the triangle. Therefore triangles of equal altitude are to each other as their bases.

## 361. - Proposition

Homologous triangles have their corresponding sides in proportion. Let the line CD (Fig. 269) be drawn parallel with. A B. Then the angles E CD and E A B are equal (Art. 331), also the angles E D C and EB A are equal. Therefore the triangles E CD and E A B are homologous, or have their corresponding angles equal.

Fig. 269.

For, join C to B, and A to D, then the triangles A C D and B CD, standing on the same base, CD, and between the same parallels, CD and A B, are equal in area. To each of these equals join the common area C DE, and the sums A DE and B CE will be equal. The triangles CD E and A D E, having the same altitude, are to each other as their bases CE and A E (Art. 360), or -

Cd E: Ad E:: Ce: A E.

Also the triangles CD E and B CE, having the same altitude, are to each other as their bases DE and BE, or -

Cde: Bce:: De: Be.

And, since the triangles A D E and B CE are equal, as before shown, therefore, substituting in the last proportion A D E for B CE, we have -

Cd E: A Be:: D E: Be.

The first two factors here being identical with the first two in the first proportion above, we have, comparing the two proportions -

Ce: Ae:: De: Be;

or, we have the corresponding sides of one triangle, CD E, in proportion to the corresponding sides of the other, A BE.

Fig. 270.

## 362. - Proposition

Two chords, E F and CD (Fig. 270), intersecting, the parallelogram or rectangle formed by the two parts of one is equal to the rectangle formed by the two parts of the other. That is, the product of C G multiplied by G D is equal to the product of E G multiplied by G F. The triangle A is similar to the triangle B, because it has corresponding angles. The angle H equals the angle G (Art. 344); the angle at J equals the angle at K, because they stand upon the same chord, DF (Art. 356); for the same reason the angle M equals the angle Z, for each stands upon the same chord, E C. Therefore, the triangle A having the same angles as the triangle B, the length of the sides of one are in like proportion as the length of the sides in the other (Art. 361). So -

Cg: Eg:: Gf: G D.

Hence, as the product of the means equals the product of the extremes (Art, 373), E G multiplied by GF is equal to C G multiplied by G D.

## 363. - Proposition

If the sides of a quadrangle are bisected, and lines drawn joining the points of bisection in the adjacent sides, these lines will form a parallelogram.

Fig. 271.

Draw the diagonals A B and CD (Fig. 271). It will here be perceived that the two triangles A E O and A C D are homologous, having like angles and proportionate sides. Two of the sides of one triangle lie coincident with the two corresponding sides of the other triangle, therefore the contained angles between these sides in each triangle are identical. By construction, these corresponding sides are proportionate; A C being equal to twice A E, and A D being equal to twice A 0; therefore the remaining sides are proportionate, CD being equal to twice E 0, hence the remaining corresponding angles are equal. Since, then, the angles A E O and A CD are equal, therefore the line E 0 is parallel with the diagonal CD - so, likewise, the line MN is parallel to the same diagonal, CD. If, therefore, these two lines, EO and MN, are parallel to the same line, CD, they must be parallel to each other. In the same manner the lines ON and EM are proved parallel to the diagonal A B, and to each other; therefore the inscribed figure ME ON is a parallelogram. It may be remarked, also, that the parallelogram so formed will contain just one half the area of the circumscribing quadrangle.