2. OF SIMILARITY OF FIGURES.

249. Figures are similar which have the same form, whether they are equal or unequal in magnitude. Figures may be similar in form, and yet be very different in magnitude; or they may be equal in magnitude, and yet be very different in form.

Figures are similar when their corresponding angles are equal, and a proportion exists between their homologous sides. By homologous sides are meant those which have the same position in the different figures with respect to the equal angles. The angles which are equal in the different figures are also called homologous angles.

Neither the equality of the angles, nor the proportion of the sides, will alone constitute similarity of figures, but the combination of the two, (fig. 135.) If the figures ABCD and EFGH are similar, then angle A = E, angle B = F, angle C = G, angle D = H, and these proportions exist:

AB: EFBC: FG
BC: FGCD:GH
CD:GH DA: HE

DA: HE AB: EF

Also, by transposition,

AB: BCEF: FG

BC: CDFG: GH

CD: DAGH: HE

DA: AB= HE: EF

In these figures angles A and E, B and F, C and G, D and H have the same position; and the sides AB and EF, BC and FG, CD and GH, DA and HE have the same position with respect to these angles.

The figures would not be similar if other angles, not

homologous, were equal, and other sides, not homologous, were proportional. If, for instance, angle A= angle E, angle B = angle G; or if the proportion were AB: EF = BC: GH.

3. TRIANGLES.

250. What ratio do the triangles ABC and DCE, (fig. 136,) which lie between the same parallels, that is, have an equal altitude, bear one to another? or in other words, How often is the surface of triangle ABC contained in the surface of triangle DCE? Or, What two other quantities are contained as often one in the other as triangle ABC is contained in triangle DCE?

Answer. Apply the base BC to the base CE as often as it can be applied, suppose 24 times, and join the division points F and G with the vertex D; then is triangle ABC = triangle DCF = triangle DFG=2 X triangle DGE. Triangle ABC is contained 24 times in triangle DCE; that is, just as often as the base BC is contained in the base CE. Thus we have triangle ABC: triangle DCE = BC : CE; that is, Triangles of equal altitude are to each other in the ratio of their bases. If 2 triangles have equal altitudes, and the base of the one is 2, 3, 4, 10, 1, times as large as the other, then is the surface of the one triangle contained 2, 3, 4, 10,, times in the surface of the other.

251. Triangles are halves of parallelograms having an equal base and an equal altitude, (152,) and since the ratio of the halves and of the whole quantities must be the same, therefore parallelograms of equal altitude are to one another as their bases.

252. (Fig. 137.) If the triangles ABC and DBC have equal bases, or a common base BC, what is the ratio of triangle ABC to triangle DBC?

Answer. Erect upon BC, at the point B, the perpendicular BF. Through the points A and D draw lines parallel to BC and intersecting BF at the points E and F; then BE is the altitude of the triangle ABC, and BF is the altitude of triangle DBC. Draw EC and FC. Triangle ABC is equivalent to EBC, and triangle DBC is equivalent to FBC, (153.) We may therefore compare together triangles EBC and FBC, instead of the triangles ABC and DBC.

Now the triangles EBC and FBC have a common altitude BC; they are to one another therefore as their bases BE and BF; thus we have the proportion,

EBC: FBC = EB : BF.

It is a principle of geometrical proportions that equal quantities may be substituted one for the other. We have therefore the proportion,

Triangle ABC: triangle DBC=EB: BF; that is, Triangles having equal bases are to each other as their altitudes.

253. Cor. Parallelograms having equal bases are to each other as their altitudes.

254. What ratio do the triangles ABC and DEF (fig. 138) bear to one another, the base and altitude of the one being respectively unequal to the base and altitude of the other?

Answer. From the vertex of the angle A let fall upon the opposite side BC the perpendicular AG; and from the vertex of the angle D let fall upon the opposite side EF the perpendicular DH.

Construct a third triangle with a base MN = BC, and an altitude NO = DH.

Let us compare together the triangles OMN and ABC, which have equal bases, and the triangles OMN and DEF, which have equal altitudes. We thus find the ratio which two quantities each bear to a third quantity, and this being known, we can find the ratio which they bear to each other.

Triangle ABC: triangle MNO = AG : ON. Triangle MNO: triangle DEF = MN: EF. Two proportions may be multiplied together, term by term, and the products will form a new proportion. Therefore triangle ABC × triangle MNO: triangle MNO X triangle DEF = AG × MN: ON X EF.

If both the terms of a ratio be divided by the same number, the ratio remains unchanged.

Therefore ABC: DEF = AG × BC: DH × EF, that is, Triangles having unequal bases and unequal altitudes, are to one another as the products of their bases multiplied by their altitudes.

255. From the preceding propositions may be deduced the general proposition, that triangles are to one another as the products of their bases multiplied by their

altitudes.

This proposition may be deduced immediately from the method of calculating the area of a triangle, (161.)

BC XAG EF × DH

:

2

triangle ABC: triangle DEF=

Or, ABC: DEF=BCXAG: EF X DH.

256. If the areas of the triangles ABC and DEF are equal, then BC × AG = EF × DH. We have here two equal products; and the factors of such products, being transposed, form a proportion.

Thus, BC: EF=DH : AG.

That is, If two triangles have equal areas, their bases are to one another in the inverse ratio of their altitudes, and the altitudes in the inverse ratio of the bases. For example, if in two triangles having equal areas, the base of one is double that of the other, then the altitude of the last must be double the altitude of the first.

257. Since triangles are the halves of parallelograms of equal bases and equal altitudes, and halves are to one another as their wholes; therefore, parallelograms are to one another as the products of their bases multiplied by their altitudes; and in parallelograms having equal areas the bases are to one another in the inverse ratio of the altitudes; and the altitudes are to one another in the inverse ratio of the bases.

Example. If the base of a parallelogram is 12 feet, and its altitude 8 feet, what is the altitude of an equivalent parallelogram, whose base is 20 feet?

Answer. 20:12 = 8:

12 × 8
20

4 feet.

258. Draw a straight line DE, intersecting two sides of the triangle ABC, and parallel to the third side BC; what ratio do the parts into which the sides are divided bear to one another, and to the entire sides? (Fig. 139.)

Sol. Draw DC and EB. The triangles DEC and EDB have a common base and an equal altitude, they are therefore equal in area. For the same reason tri