PROPOSITION X. THEOREM.

If the two sides AB, AC, of the triangle ABC, are equal respectively to the two sides DE, DF, of the triangle DEF, while, at the same time, the angle BAC contained by the former, is greater than EDF contained by the latter; then will the third side, BC of the first triangle, be greater than the third EF of the second.

MAKE the angleCAG =D, take AG-DE, and join CG. The triangle GAC is equal to DEF (Prop. 6.), since, by construction, they have an equal angle in Beach, contained by equal sides; therefore

CG is equal to EF.

C

E

D

Now, there may be three cases in the proposition, according as the point G falls without the triangle ABC, or upon its base BC, or within it.

First Case. The straight line GC is shorter than GI+IC, and the straight line AB is shorter than AI+IB; therefore, GC+AB is shorter than GI+AI+IC+IB, or, which is the same thing, GC+ABAG+BC. Take away AB from the one side, and its equal AG from the other, there will remain GCBC; but we have found GC=EF, therefore, EF

Second Case. If the point G fall on the side BC, it is evident that GC, or its equal EF, will be shorter than BC.

BC.

Third Case. Lastly, if the point G fall within the triangle ABC, we shall have, by the preceding theorem, AG+ GCZAB+BC; and, taking AG from the one side, and its equal AB from the other, there will remain GC BC, or B EF BC.

Scholium. Conversely, if the two sides AB, AC of the triangle ABC are equal to the two DE, EF of the triangle DEF, while the third side, CB of the first triangle, is greater than the third EF of the second; then will the angle BAC of the first triangle be greater than the angle EDF of the second.

E

For if not, the angle BAC must be equal to EDF, or less than it. In the first case, the side CB would (Prop. 6.) be equal to EF; in the second, CB would be less than EF: but both of these results contradict the hypothesis; therefore, BAC is greater than EDF.

PROPOSITION XI. THEOREM.

Two triangles are equal, when the three sides of the one are respectively equal to the three sides of the other.

LET the side AB (see the diagram of Prop. 6.) be equal to DE, AC=DF, BC EF; then is the angle Á=D, B=E, C-F.

For, if the angle A were greater than D, since the sides AB, AC are respectively equal to DE, DF, it would follow, by the last proposition, that the side BC must be greater than EF: and, if the angle A were less than D, it would follow that the side BC must be less than EF. But BC is equal to EF; therefore the angle A can neither be greater nor less than D ; therefore they are equal. In the same manner it may be shewn, that B is equal to E, and C to F.

Scholium. It may be observed, that the equal angles lie opposite to the equal sides: thus the equal angles A and D lie opposite to the equal sides BC and EF

PROPOSITION XII. THEOREM.

In an isosceles triangle, the angles opposite the equal sides are equal.

LET the side AB be equal to AC, the angle C will be equal to B.

Join A the vertex, and D the middle point of the base BC. The triangles ADB, ADC, have all the sides of the one respectively equal to those of the other, AD being common, AB=AC (hyp.) and BD DC by construction; therefore, by the last proposition, the angle B is equal to the angle C.

B

D

C

Cor. An equilateral triangle is likewise equiangular, that is to say, has all its angles equal.

Scholium. The equality of the triangles ABD, ACD, proves also that of the angle BAD to DAC, and of BDA to ADC; hence the latter two are right angles; hence the line drawn from the vertex of an isoceles triangle to the middle point of its base, is perpendicular to that base, and divides the angle at the vertex into two equal parts.

In a triangle which is not isosceles, any side may be assumed indifferently as the base; and the vertex is, in that case, the vertex of the opposite angle. In an isosceles triangle, however, that side is specially assumed as the base, which is not equal to either of the other two.

PROPOSITION XIII. THEOREM.

Conversely, if two angles of a triangle are equal, the sides opposite them will be equal, and the triangle will be isosceles.

LET the angle ABC be equal to ACB; then will the side AC be equal to the side AB.

For, if those sides are not equal, let AB be the greater. Take_BD=AC, and join DC. The angle DBC is (Hyp.) equal to ACB; the two sides DB, BC, are equal to the two AC, CB, by construction; therefore (Prop. 6.), the triangle DBC must be equal to ACB. But the part cannot be equal to the whole; hence there is no inequality between the sides AB, AC; hence the triangle ABC is isosceles.

B

PROPOSITION XIV. THEOREM.

Of two sides of a triangle, the greater is that which lies opposite the greater angle; and conversely, of two angles of a triangle, the greater is that which lies opposite the greater side.

First, Let the angle C be greater than B; then will the side AB, opposite to C, be greater than AC, opposite to B.

B

Make the angle BCD=B. Then in the triangle BDC, we shall have BD=DC. (Prop. 13.) But the line AC is shorter than AD+DC, and AD+ DC=AD+DB=AB; therefore, AB is greater than AC. Secondly. Suppose the side AB AČ; then will the angle C, opposite to AB, be greater than the angle B, opposite to AC. For, if we had CB, it would follow, from what has just been proved, that we must have ABAC, which is contrary to the hypothesis. If we had C=B, it would follow (Prop. 13.) that we must have AB-AC, which is also contrary to the hypothesis. Therefore, the angle C must be greater than B.

PROPOSITION XV. THEOREM.

From a given point A without the straight line DE, no more than one perpendicular can be drawn to that line.

FOR, suppose we can draw two, AB and

AC.

Produce one of them AB, till BF is

equal to AB, and join FC.

A

B E

The triangle CBF is equal ABC; for the angles CBF and CBA are right, the D side CB is common, and the side BF=AB: therefore (Prop. 6.), those triangles are equal, and the angle BCF-BCA. The angle BCA is right, by hypothesis; therefore BCF must be right also. But if the adjacent angles BCA, BCF are together equal to two right angles, the line ACF must (Prop. 4.) be straight; from whence it follows, that between the same two points A and F, two straight lines can be drawn; which is impossible: hence it is equally impossible that two perpendiculars can be drawn from the same point, to the same straight line. Scholium. At one given point C, in the line AB, it is equally impossible to erect two perpendiculars to that line; for (see the diagram of Prop. 2.), if CD and CE were those two perpendicu

lars, the angle DCB would be right, as well as BCE, and the part would thus be equal to the whole.

PROPOSITION XVI. THEOREM.

If from a point A (see the preceding figure), situated without the straight line DE, a perpendicular AB be let fall on that straight line, and several oblique lines AE, AC, AD, be drawn to several points in the same line.

First, The perpendicular AB will be shorter than any oblique

line.

Secondly, The two oblique lines, AC, AE drawn on different sides of the perpendicular, at equal distances BC, BE, will be equal.

Thirdly, Of two oblique lines, AC, AD, or AE, AD, drawn at pleasure, the one which lies farther from the perpendicular will be the longer.

Produce the perpendicular AB till BF is equal to AB, and join FC, FD.

First. The triangle BCF, is equal to the triangle BCA, for they have the right angle CBF-CBA, the side CB common, and the side BF=BA; hence the third sides, CF and AC are equal. But ABF, being a straight line, is shorter than ACF, which is a broken line; therefore AB, the half of ABF, is shorter than AC, the half of ACF: therefore the perpendicular is shorter than any oblique line.

Secondly. If we suppose BE-BC; since we have, farther, the side AB common, and the angle ABE=ABC, the triangle ABE must be equal to the triangle ABC; hence the sides AE, AC are equal; hence two oblique lines equally distant from the perpendicular are equal.

Thirdly. In the triangle DFA, the sum of the lines AC, CF, is less (Prop. 9.) than the sum of the sides AD, DF; therefore AC, the half of the line ACF, is shorter than AD, the half of the line ADF; therefore such oblique lines as lie farthest from the perpendicular are longest.

Cor. 1. The perpendicular measures the true distance of a point from a line, because it is shorter than any other distance.

Cor. 2. From the same point three equal straight lines cannot be drawn to the same straight line; for if there could, we should have two equal oblique lines on the same side of the perpendicular, which is impossible.