greater than GC. But we have found that GC is equal to EF; therefore, BC will be greater than EF.

of the first is greater than the third side EF of the second, then the angle BAC of the first triangle will be greater than the angle EDF of the second.

For, if not greater, the angle BAC must be equal to EDF or less than it. In the first case, the side BC would be equal to EF (P. 5, c), in the second, BC would be less than EF; but either of these results contradicts the hypothesis therefore, BAC is greater than EDF

PROPOSITION X. THEOREM.

If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are equal.

Let EDF and BAC be two triangles, having the side ED=BA, the side EF-BC, and the side DF-AC; then will the angle D=A, the angle E=B, and the angle F=C, and consequently the triangle EDF will be equal to the triangle BAC.

that the side EF would be greater than BC; and if the angle D were less than A, the side EF would be less than BC. But EF is equal to BC, by hypothesis; therefore, the angle D can neither be greater nor less than A; therefore it must be equal to it. In the same manner it may be shown that the angle E is equal to B, and the angle F to C: hence, the two triangles are equal (P. 6, s).

Scholium. It may be observed, that when two triangles are equal to each other, the equal angles lie opposite the equal sides, and consequently, the equal sides opposite the equal angles: thus, the equal angles D and A, lie opposite the equal sides EF and BC.

PROPOSITION XI. THEOREM.

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

equal.

Let BAC be an isosceles triangle, having the side BA equal to the side AC; then will the angle C be equal to the angle B.

For, join the vertex A, and the middle point D, of the base BC. Then, the triangles BAD, DAC, will have all the sides of the one equal to those of the other, each to each. For, BA is equal to

AC, by hypothesis, AD is common, and

A

B

D

BD is equal to DC by construction: therefore, by the last proposition, the angle B is equal to the angle C.

Cor. 1. An equilateral triangle is likewise equiangular, that is to say, has all its angles equal.

Cor. 2. The equality of the triangles BAD, DAC, proves also that the angle BAD, is equal to DAC, and BDA to

ADC; hence, the latter two are right angles. Therefore, the line drawn from the vertex of an isosceles triangle to the middle point of the base, divides the angle at the vertex into two equal parts, and is perpendicular to the base.

Scholium. 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 generally assumed as the base, which is not equal to either of the other two.

PROPOSITION XII. THEOREM.

Conversely: If two angles of a triangle are equal, the sides opposite them are also equal, or, the triangle is isosceles.

In the triangle BAC, let the angle B be equal to the angle ACB; then will the side AC be equal to the side AB.

B

A

D

For, if these sides are not equal, suppose AB to be the greater. Then, take BD equal to AC, and draw CD. Now, in the two triangles BDC, BAC, we have BD=AC, by construction; the angle B equal to the angle ACB, by hypothesis; and the side BC common: therefore, the two triangles, BDC, BAC, have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each: hence they are equal (P. 5). But the part cannot be equal to the whole (A. 8); hence, there is no inequality between the sides BA and AC; therefore, the triangle BAC is isosceles.

PROPOSITION XIII. THEOREM.

The greater side of every triangle is opposite to the greater angle; and conversely, the greater angle is opposite to the greater side.

First, In the triangle CAB, let the angle C be greater than the angle B; then will the side AB, opposite C, be greater than AC, opposite B.

Secondly. Suppose the side AB>AC; then will the angle C, opposite to AB, be greater than the angle B, opposite to AC.

For, if the angle CB, it follows, from what has just been proved, that AB <AC; which is contrary to the hypothesis. If the angle C-B, then the side AB-AC (P. 12); which is also contrary to the supposition. Therefore, when AB>AC, the angle C cannot be less than B, nor equal to it; therefore, the angle must be greater than B.

PROPOSITION XIV. THEOREM.

From a given point, without a straight line, only one perpendicular can be drawn to that line.

A

D

-E

B

F

Let A be the point, and DE the given line. Let us suppose that we can draw two perpendiculars, AB, AC. Prolong either of them, as AB, till BF is equal to AB, and draw FC. Then the two triangles CAB, CBF, will be equal for, the angles CBA and CBF are right angles, the side CB is common, and the side AB equal to BF, by construction; therefore, the two triangles are equal, and the angle ACB=BCF (P. 5, c). But the angle ACB is a right angle, by hypothesis; therefore, BCF must likewise be a right angle. Now, if the adjacent angles BCA, BCF, are together equal to two right angles, ACF must be a straight line (P. 3). Whence, it follows, that between the same two points, A and F, two straight lines can be drawn, which is impossible (A. 11): therefore, only one

perpendicular can be drawn from the same point to the same straight line.

Cor. At a given point C, in the line AB, it is equally impossible to erect more than one perpendicular to that line. For, if CD, CE, were both perpendicular to AB, the angles A BCD, BCE, would both be right

E

D

V

-B

angles; hence, they would be equal (A. 10), and a part would be equal to the whole, which is impossible.

PROPOSITION XV. THEOREM.

If from a point without a straight line, a perpendicular be let fall on the line, and oblique lines be drawn to different points:

1st. The perpendicular will be shorter than any oblique line. 2d. Any two oblique lines which intersect the given line at points equally distant from the foot of the perpendicular, will be equal.

3d. Of two oblique lines which intersect the given line at points unequally distant from the perpendicular, the one which cuts off the greater distance will be the longer.

Let A be the given point, DE the given line, AB the perpendicular, and AD, AC, AE, the oblique lines.

Prolong the perpendicular AB till BF is equal to AB, and draw FC, FD.

First. The triangle BCF, is equal to the triangle CAB, for they have the right angle CBF=CBA, the side CB common, and the side BF=BA; hence, the third sides, CF and CA

A

D

E

B

F

are equal (P. 5, c). But ABF, being a straight line, is shorter than ACF, which is a broken line (A. 12); therefore, AB, the half of ABF, is shorter than AC, the half of ACF; hence, the perpendicular is shorter than any oblique line.