but the figure KAB is to the figure LCB as the figure MF is to the figure NH,

Hyp.

therefore the duplicate ratio of AB : CD is equal to the duplicate

ratio of EF: GH.

Therefore AB: CD:: EF: GH.

IV. 14, Part ii.

Q.E.D.

THEOR. 14. If two triangles or parallelograms have one angle of the one equal to one angle of the other, their areas have to one another the ratio compounded of the ratios of the including sides of the first to the including sides of the second.

Let ABC, DBE be two triangles having the angle ABC equal to the angle DBE :

then shall the triangle ABC have to the triangle DBE the ratio compounded of the ratios BC: BE and BA : BD.

Apply the triangle ABC to the triangle DBE,

so that BC may fall along BE,

then BA will fall along BD,

since the angle ABC is equal to the angle DBE. Join AE.

Hyp.

Then the triangle ABC has to the triangle DBE the ratio compounded of the ratios of the triangle ABC to the triangle ABE, and of the triangle ABE to the triangle DBE.

But the ratio of the triangle ABC to the triangle ABE is equal to the ratio BC BE, IV. 12, Cor. and the ratio of the triangle ABE to the triangle DBE is equal to the ratio AB : DB. IV. 12, Cor. Hence the ratio of the triangle ABC to the triangle DBE is equal to the ratio compounded of the ratios BC: BE and BA : BD.

A like proof may be given for parallelograms.

Q.E.D.

COR. I. If two triangles or parallelograms have one angle of the one supplementary to one angle of the other, their areas have to one another the ratio compounded of the ratios of the including sides of the first to the including sides of the second.

COR. 2. The ratio compounded of two ratios between straight lines is the same as the ratio of the rectangle contained by the antecedents to the rectangle contained by the consequents.

Ex. 42. D is any point in AC, the base of an isosceles triangle ABC. DE, DF make equal angles with AC, and meet the equal sides BC and AB in E and F. Prove that the triangles AED and CDF are equal.

Ex. 43. If ABC is a triangle right-angled at B, and BD the perpendicular on AC is produced to E, so that DE is a third proportional to BD and DC, the triangle ADE will be equal to the triangle BDC.

THEOR. 15. Triangles and parallelograms have to one another the ratio compounded of the ratios of their bases and of their altitudes.

Let ABC, HKL be two triangles, AD, HM their altitudes:

E

B

H

K

L M

then shall the triangle ABC be to the triangle HKL in the ratio compounded of the ratios BC: KL and AD : HM.

From B draw BE at right angles to BC, and equal to AD, join EC;

and from K draw KN at right angles to KL and equal to HM, join NL.

II. 2. Cor. I.

Then the triangles ABC and EBC are equal, and so also are the triangles HKL and NKL. But because the angle EBC is equal to the angle NKL, therefore the triangle EBC is to the triangle NKL in the ratio compounded of the ratios BC: KL and EB: NK,

that is, of the ratios BC: KL and AD: HM.

V. 14.

Therefore the triangle ABC is to the triangle HKL in the ratio compounded of the ratios BC: KL and AD: HM.

A like proof may be given for parallelograms.

Q.E.D.

Ex. 44. ABCD is a quadrilateral, having the side AB parallel to CD. The diagonals meet in O. Prove that the rectangle AO, OD is equal to the rectangle BO, OC.

THEOR. 16. In a right-angled triangle, any rectilineal figure described on the hypotenuse is equal to the sum of two similar and similarly described figures on the sides.

Let ABC be a triangle having the angle BAC a right angle, and let H, K, L be similar and similarly described rectilineal figures on BC, CA, AB :

B

H

K

then shall the figure H be equal to the sum of the figures K

and L.

Draw AD perpendicular to BC.

Then BC CA:: CA: CD,

V. 7. Cor.

V. 12.

therefore BC: CD is the duplicate of the ratio BC : CA, therefore the figure K is to the figure H as CD to BC. In like manner the figure L is to the figure H as BD to BC. Therefore the sum of the figures K and L is to the figure H as the sum of CD and BD is to BC.

IV. 8.

But the sum of CD and BD is equal to BC, therefore the sum of the figures K and L is equal to the figure H.

Q.E.D.

Ex. 45. If these figures are rectangles, prove the proposition by a method analogous to that used in the first proof of II. 9.

THEOR. 17. The rectangle contained by the diagonals of a quadrilateral is less than the sum of the rectangles contained by opposite sides unless a circle can be circumscribed about the quadrilateral, in which case it is equal to that sum.

Let ABCD be a quadrilateral about which a circle cannot be described:

E

B

then shall the rectangle contained by AC, BD be less than the sum of the rectangles contained by AB, CD and by AD, BC.