Let A C be produced and let BD be a perpendicular on the prolongation of A C; then C D is the projection of B C on A D [I. Def. 19]. Because A B is the hypothenuse of the right-angled triangle ABD, the square of A B is equivalent to the squares of B D and AD [11]: but the square of A D is equivalent to the squares of AC and CD, together with twice the rectangle of AC by CD [8]; therefore the square of AB is equivalent to the squares of BD, CD and AC together with twice the rectangle of A C by CD. Now the squares of BD and C D are together equivalent to that of BC [11]; therefore the square of AB is equivalent to the squares of BC and AC together with twice the rectangle of A C, by the projection of B C on the prolongation of A C, which W. T. B. D.

COROLLARY. The square of the side subtending an acute angle of a triangle, is equivalent to the difference between the squares of the other sides, and twice the rectangle of one of the latter by the projection thereon of the other. (Eucl. II. 13.)

THEOREM 13.

The squares of two sides of a triangle, are together equivalent to twice the square of half the other side, with twice that of the median thereto.

A

E

Let ABC be a triangle, and CD the median to the side A B.

Let C E be a perpendicular to A B; then D E is the projection of CD on A B. Because the square of A C is equivalent to those of AD and CD, together with twice the rectangle of AD by DE [12]; and because the square of CB with twice the rectangle of D B by D E, is equivalent to the squares of D B and C D [12]; the squares of A C and CB with twice the rectangle of AD by D E, are together equivalent to the squares of AD and D B with twice that of CD, and twice the rectangle of DB by DE: but, because A D is equal to DB [Const.], the square of AD is equivalent to that of DB [III. 39 i], and the rectangle of AD by D E is equivalent to the rectangle of D B by D E [III. 39 ii]; therefore the squares of A C and C B are together equivalent to twice those of AD and CD [iv], which W. T. B. D.

Scholium. The squares of the equal sides of an isosceles triangle, are together equivalent to twice the square of half the base, with twice that of the altitude.

THEOREM 14.

The squares of the sides of a quadrangle, are together equivalent to those of its diagonals, with four times that of the middle line of the diagonals.

Let there be a quadrangle A B C D and let E F be the middle line of its diagonals A C and BD.

Let A F and F C be joined; then F E is a median of the triangle AFC. Because the squares of A B and AD are together equivalent to twice those of D F and A F [12], and because the squares of BC and CD are together equivalent to twice those of D F and F C, the squares of AB, A D, BC, and CD are together equivalent to four times the square of D F with twice those of AF and FC [vi]. Because the squares of A F and F C are together equivalent to twice those of A E and E F, twice the squares of AF and F C are equivalent to four times those of A E and EF [vi]; therefore the squares of AB, AD, B C and CD are together equivalent to four

times those of DF, AE and EF: but four times the squares of DF and A É are equivalent to the squares of DB and AC [8 i];

B

D

therefore the squares of the sides of the quadrangle ABCD, are together equivalent to those of the diagonals D B and A C, with four times that of their middle line E F, which W. T. B. D.

COROLLARY. The squares of the sides of a parallelogram, are together equivalent to those of its diagonals.

SECTION III.

ON THE COMPARISON OF THE SIDES OF EQUIVALENT PLANE
FIGURES.

THEOREM 15.

(Eucl. VI. 1.)

Two parallelograms having equal altitudes are as their bases.1

Let ABCD and EFHI be two parallelograms having equal altitudes.

Let the bases AB and EF be divided into equal parts small enough to be contained an equal number of times in each one; let, for instance, A B contain seven, and E F three of these parts: then A B is to E Fas seven is to three [x].

Now, let there be drawn, through each point of division of A B and EF, a parallel to the adjacent sides. Then the two parallelograms A C, EH, are divided into parallelograms having all equal bases and altitudes, and being consequently all equivalent to one another [1]; but

1In The Elements, the expression that two or more figures are as two or more others, is used with the meaning that the first are proportional to the others.

AC will contain seven, and EH three of these parallelograms; then the parallelogram A C is to the parallelogram EH as seven is to three [x].

Therefore, the parallelogram AC is to the parallelogram E H as the base A B is to the base E F [i].

Should the lines A B and E F be incommensurable, let them be imagined to be divided into their infinitely small elements, and let a parallel to the adjacent sides be imagined through each point of division; then the two parallelograms will be divided into a number, finite or infinite, of parallelograms having equal bases [Post. IV] and altitudes [Hyp.]; therefore these partial parallelograms will be equivalent to one another [1]. However small these partial parallelograms may be, it is evident that the number of them in each of the parallelograms AD and EH, will be the same as that of the parts into which A B and EF have been divided; therefore the two parallelograms are still to each other as their bases W. W. T. B. D.

COROLLARY I. Two triangles having equal altitudes are as their

bases.

COROLLARY II. Two parallelograms, or triangles, which are between the same parallels, are as their sides thereon.

COROLLARY III. Two parallelograms, or triangles, having equal bases, are as their altitudes.

COROLLARY IV. The rectangles of a straight line by two others are as the latter.

COROLLARY V. Two parallelograms having equal altitudes, are as the rectangles of their bases by the same straight line.

THEOREM 16.

(Eucl. VI. 14.)

The sides of the homologous angles of two equivalent equiangular parallelograms, are reciprocally proportional.

Let there be two equiangular parallelograms ABCD and AEFH