angle EBC = DBC. equals; therefore DEC + ADE = ADC = EDB + ADE=AЕВ.

Equals added to equals give

Equal quantities bear an equal ratio to equal or the same third quantities; or, a third quantity bears the same ratio to two equal quantities.

Thus ADE : DEC=ADE: EDB

But ADE: DEC = AE : EC (250)

And ADE: EDB = AD: DB (250)

Therefore AE : EC = AD : DB; that is, the parts of one side are proportional to the parts of the other side, directly.

Again, triangle ADC = AEB, therefore
ADC: ADE=AEB: ADE.

From this we may deduce the proportion,
AC:AE AB : AD.

And, because triangle ADC = AEB and triangle DEC = EDB, therefore

ADC: DEC=AEB: EDB;

from which we may deduce the proportion,

AC: EC = AB : DB.

That is, Each entire side is to one of its parts, as the other entire side is to its corresponding part.

259. By the transposition of the middle terms of a proportion, a new proportion may be formed; and thus in the figure under examination many more proportions may be proved to exist :

Thus, AB: AD=AC: AE

AB: AC = AD : AE

AB: BD = AC : EC

AB: AC=DB : EC

BD: DA - CE: EA

BD:CE=DA : ΕΛ

260. We will now seek the ratio which the base of the triangle bears to the parallel line intersecting the sides.

(Fig. 140.) Draw EG parallel to AB. From the preceding proposition we get the following proportion, CB: CA = GB: EA

Now GB=ED, being opposite sides of a parallelogram, therefore CB:CA= ED: EA

by transposition CB: ED=CA: EA

But CA: EA = BA : DA (258)

therefore CB:ED=BA: DA

that is, the base of a triangle is to the parallel line intercepted by the other sides, as one of those sides is to that part of it cut off by the parallel which lies next to the vertex of the triangle.

261. To find a fourth proportional line to three given lines.

Sol. Let M, N, and O be the given lines, (fig. 141.) Draw two indefinite straight lines making any angle with each other. In one of them take AB=M, and BC=N. In the other take AD=0. Draw BD, and through C draw CE parallel to BD. DE is the line required.

For, BD is parallel to CE,

therefore AB: BC=AD: DE
that is,
M:NO: DE

262. Remark 1. If N and O are equal, the required line will be a third proportional to the lines M and N. 2. If M and N are equal, O will be equal to the required line.

263. If the sides of an angle are divided proportionally, the straight lines joining the division points will be parallel.

(Fig. 141.) If AC: AB=AE: AD

then BD is parallel to CE.

Dem. If BD is not parallel to CE, then a line may

be drawn which shall be parallel; suppose it to be BX.

then AC: AB=AE: AX

But by supposition AC: AB=AE: AD therefore AD = AX, which is impossible. Therefore BD is parallel to CE.

Conditions which determine the similarity of two Triangles. 264. Two triangles are similar, if two angles of the one are equal to two angles of the other, each to each.

Dem. Let angle B = E, and C = F, (fig. 142.) Since two angles of one triangle are equal to two angles of the other, the third angle of the one must be equal to the third angle of the other, (123.) Consequently all the angles of the one are equal to the angles of the other, each to each.

To ascertain the proportionality of the sides, let us suppose the triangle ABC to be placed upon DEF, so that AB shall fall on DE, AC on DF, and BC on GH. Triangle ABC = DGH, (137;) consequently angle ABC=DGH, and angle ACB=DHG. But angle ABC=DEF, therefore DGH = DEF,) 112;) therefore GH is parallel to EF.

Thus the homologous sides of the two triangles are proportional, and since the homologous angles are equal, the triangles are similar.

265. Two triangles are similar, if the sides of one are proportional to the homologous sides of the other, each to each, and the included angles are equal.

Dem. (Fig. 142.) Suppose AB: DE = AC : DF and angle A = D. Place the triangle ABC upon EDF so that point B shall fall on G, and point C on H. Then is triangle ABC = DGH, (126;) therefore BC = GH, angle ABC=DGH and ACB=DHG.

By supposition AB:DE=AC: DF therefore DG:DE=DH : DF

therefore GH is parallel to EF; therefore angle DGH = DEF and angle DHG = DFE, (108;) consequently triangle DGHS DEF; therefore triangle ABCS DEF.

266. Remark. The sides which are opposite to equal angles are proportional; and the angles which are opposite to proportional sides are equal.

267. Two triangles ABC and DEF (fig. 143) are similar, if the three sides of the one are proportional to the homologous sides of the other, each to each.

Dem. Since the sides of the two triangles are by supposition proportional, we have to seek only for the equality of the homologous angles. It will be sufficient to demonstrate that one angle of the one triangle is equal to the homologous angle of the other, for then, by the preceding proposition, the triangles will be similar.

Upon EF at the point E make an angle FEG=ABC; and at the point F make an angle EFG=ACB. Tri

angles ABC and GEF are similar, (264;) therefore angle A G. We have therefore the proportion;

Now EF = EF, therefore triangle GEF = DEF; therefore angle G=D. But angle G = A; therefore A =D. Consequently ABC S DEF.

268. Remark. In the triangles ABC and DEF the proportional sides are opposite to the equal angles.

269. From the preceding propositions we learn that the similarity of two triangles may be inferred;

1. If two angles of the one are equal to the homologous angles of the other, each to each.

2. If two sides of the one are proportional to the homologous sides of the other, and the included angles equal.

3. If the three sides of the one are proportional to the homologous sides of the other.

We also learn that in similar triangles the equal angles are opposite to the proportional sides, and the proportional sides to the equal angles.

4. MISCELLANEOUS PROPOSITIONS.

270. To construct upon the base BC a triangle which shall be similar to a given triangle, DEF, (fig. 143.)