Draw PFH parallel to AB, intersecting AC at F; and upon AF construct the parallelogram AFHI, equal in area to the given area of the triangle. Make IK perpendicular to AI, and equal to FP; and, from the point K to AB, apply KD = PH; then draw DPE, and the problem will be solved.

For, supposing M to be the intersection of DE and IH, it is evident, because of the parallel lines, that the three triangles PHM, PFE and MDI are equiangular. Therefore, all equiangular triangles being in the ratio of the squares of their homologous sides (B. IV, Prop. xi), and the sum of the squares of FP or IK and DI being equal to the square of PH or KD by construction (B. II, Prop. vIII), it is evident that the sum of the triangles PFE and DMI is equal to the triangle PHM (this is obvious also from B. IV, Prop. xvi, Cor.); to which equal quantities in fig. 1, add AFPMI, and we shall have ADE equal to AFHI. But in fig. 2, let PFE be taken from PHM, and there will remain EFHM = DMI; to each of which, adding AIME, we have AFHI ADE as before.

PROPOSITION XVII.

PROBLEM. On a given line GH homologous to a given side AB of a given rectilineal figure ABCDF, to construct a figure similar to the given one.

From one of the extremities of the line AB which is homologous to GH, draw lines to all the angles of the figure; draw HK, making the angle GHK equal to the

angle ABC (B. I, Prop. x); also draw GK, making the angle HGK equal to the angle BAC; then will the triangle GHK be similar to the triangle ABC (B. IV, Prop. VII). In a similar manner, on GK, homologous with AC, construct the triangle KGL similar to the triangle ACD; also on the side GL, homologous with AD, construct the triangle GLM similar to the triangle ADF. Then will the polygon GHKLM be similar to the polygon ABCDF.

For, these two polygons are composed of the same number of similar triangles similarly situated, and therefore they are similar (B. IV, Prop. xv, Schol.).

(75.) Problem. To make a polygon similar to a given polygon, and having its perimeter in a given ratio to the perimeter of the given polygon.

Let ABCDF be the given polygon.

From any point either within or without the polygon, as G, draw lines to all the angles of the polygon. Then take GH, GK, GL, GM, GN, each in the same ratio to the corresponding lines GA, GB, GC, GD, GF, that the perimeter of the required polygon is to be to the perimeter of the given polygon. Join the points H, K, L, M, N, and the polygon HKLMN will be the polygon required.

These polygons are evidently similar, for the corresponding sides are parallel (B. IV, Prop. 11); and since, in the triangle GAB, HK

is parallel to AB, we have HK: AB :: GK : GB; and as the side

KL is parallel to BC, we have GK GB:: KL BC. Therefore, by equality of ratios, we have HK : KL :: AB: BC.

In the same way it may be shown that KL : LM :: BC: CD, and so on for the other sides. Therefore these polygons are similar.

(76.) PROBLEM. To draw a complex figure similar to another figure, on the same or different scales, by means of rectangles.

Surround the given figure by a square or a rectangle of convenient size, and divide it by pencil lines, intersecting perpendicularly, into squares or rectangles, as small as may be deemed necessary. Generally, the more irregular the contour of the figure, or the more numerous the sinuosities or subdivisions, the more numerous the rectangles should be..

Then draw another square or rectangle, having its sides either equal to the former, or greater or less in the assigned ratio, and divide this figure into as many squares or rectangles as there are in the original figure. Draw, in every rectangle of the new figure, right lines or curved, to agree with what is contained in the corresponding rectangle of the original figure; and this, if carefully done, will give a correct copy of the complex diagram proposed.

In ornamental needle-work, the same system of copying is practised. The figures to be executed are usually required to be wrought on coarse canvas, the threads of which form a system of squares, such as above described. The original object from which the copy is made, is delineated in proper colors on paper on which a similar system of squares is printed, the color occupying each square being there distinctly expressed; so that the needle-worker sees, at once, what color of silk or worsted it is necessary to make use of for each respective square.

(77.) The Proportional Compass is an instrument much used by artists in the reduction of pictures, etc. It consists of two similar and equal pieces of brass AB and CD, terminated at each end by steel points. F is a pivot, which may be adjusted so as to divide the length of the two legs into any required ratio. Whatever be the ratio of AF to FB, or of CF to FD, the same will be the ratio of the distance AC to BD, whatever may be the extent to which the compass is opened.

B

PROPOSITION XVIII.

THEOREM. In a right-angled triangle, a perpendicular drawn from the right angle to the hypothenuse is a mean proportional between the segments of the hypothenuse ; and each of the sides about the right angle is a mean proportional between the hypothenuse and adjacent seg

ment.

Let ABC be a rightangled triangle, and CD a perpendicular from the right angle C to the hypothenuse AB; then will

or

AC

CD be a mean proportional between AD and DB,
a mean proportional between AB and AD, and
a mean proportional between AB and BD;

BC

For, the two triangles ABC, ADC, having the right angles at C and D equal, and the angle A common, have their third angles equal, and are equiangular (B. I, Prop. XXIV, Cor. 1). In like manner, the two triangles ABC, BDC, having the right angles at C and D equal, and the angle B common, have their third angles equal, and are equiangular. Hence all the three triangles ABC, ADC, BDC, being equiangular, will have their like sides proportional (B. IV, Prop. ví), viz.

Cor. 1. Since an angle in a semicircle is a right angle (B. III, Prop. vIII, Cor. 3), it follows, that if, from any point C in a semicircumference, a perpendicular be drawn to the diameter AB, and the two chords CA, CB be drawn to the extremities of the diameter, then are CD, AC, BC the mean proportionals as in this proposition; or

CD' AD.DB, AC' AB. AD, and BC=AB. BD.

=

Cor. 2. Hence AC2: BC2 :: AD: BD.

Cor. 3. Hence we have another demonstration of Proposition VIII, Book II.

For, since AC2=AB. AD, and BC'=AB.BD; by addition, AC+BC' = AB (AD+BD) = AB'.

PROPOSITION XIX.

PROBLEM. To find a mean proportional between two given lines AB, BC.