SOME IMMEDIATE DEVELOPMENTS OF THE PROPS. IN BOOK vi.--NOT SO OBVIOUS AS TO BE PROPERLY CALLED COROLLARIES.

THEOREM (1) Triangles (or parallelograms) are to each other in the ratio compounded of the ratios of their bases and altitudes.

Let X, X' be the areas of two As (or two s);

a, a' their respective altitudes;

b, b' their corresponding bases.

Then, if Y is the area of a ▲ (or □) of altitude a, and base b',

Cor. (1). If triangles (or parallelograms) have equal areas, any pair of their altitudes are reciprocally proportional to the bases to which they are drawn. Cor. (2). Triangles (or parallelograms) have the same ratio to each other as rectangles under their respective altitudes and bases.

Cor. (3). Since in equiangular triangles, the altitudes are as the bases to which they are drawn, vi. 19 is an immediate deduction from the above. Cor. (4). Similarly vi. 23 is deducible from it.

THEOREM (2)—If two lines are cut by three parallels, the intercepts on the one are in the same ratio as the corresponding intercepts on the other.

Let the three Is AA', BB', CC', cut other two lines in A, B, C and A', B',

C' respectively.

Join AB', A'B, BC', B'C.

Note This Theorem (which is here deduced from vi. 1) might have been proved directly from the definition of proportion, in the same way as vi. I.

THEOREM (3)—If two similar unequal rectilineal figures are so placed that their corresponding sides are parallel, then the joins of corresponding corners are all concurrent.

.. O and O' are the same pt.

i. e. AP, BQ, CR are concurrent.

Similarly all corresponding joins are concurrent in O.

Def. The point so determined is called a centre of similarity of the figures. Cf. p. 253.

Note-Figures like the above are said to be similarly situated with respect to each other.

Extension of vi. 13—Philo's mode of finding two mean proportionals between two lines, by the use of a graduated ruler.

Place the lines in positions AB, AC, at rt. ^3.

Complete rect. ABPC; and describe O round this rect. Now place a graduated ruler, with its edge at P, and meeting AB, duced respectively in X, Y; and the O in Q.

AC pro

Note-A line XPQY (drawn as in fig.) through any pt. P, within any angle A; so that, AQ being perpendicular to XY, then QX, PY are equal, is called Philo's Line: it possesses the property, the proof of which is difficult (see Mulcahy's Modern Geometry, p. 106), that XPY is the least line through P terminated by the sides of the angle.

THEOREM (4)-(Another Converse to the first part of vi. 14 or 15). If two parallelograms (or triangles) have equal areas, and the sides about a pair of angles reciprocally proportional, then the angles contained by these sides are either equal, or supplementary.

This may be proved by drawing the altitudes to an homologous pair of the reciprocal sides; and then using vi. Addenda (1) Cor. (1), and vi. 7.

THEOREM (5)-(Converse of vi. 23). If parallelograms have to each other the ratio which is compounded of the ratios of their sides, then they are equiangular to each other.

Assume that H, G are respectively at X, Y, which do not coincide with P, R.

Ĝa

rt. A.

of BPX and BXP, one a rt. A, and one rt. ^.

And they cannot each be right.

.. they are unequal.

But they were before proved equal.

the assumption that X, Y do not coincide with P, R has led to a contradiction; and .. is not true.

THEOREM (6)—If the hypotenuse of a right-angled triangle is divided in extreme and mean ratio by the altitude drawn to it; then—

(a) the lesser side containing the right angle is equal to the alternate segment of the hypotenuse; and conversely:

(B) the greater side containing the right angle is a mean proportional between the hypotenuse and the remaining side; and conversely.

Cor. From the above, a right-angled triangle can be constructed on a given line as hypotenuse, and so that its sides are in continued proportion: for if AB is the given line, then dividing it in N in extreme and mean ratio, and drawing NC perpendicular to it to meet semi-circle on AB in C, gives ABC such a triangle.