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


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

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

For, if these sites 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;B and the side BC common: therefore, the two triangles, BDC, BAC, have two sides and the included angle in the one, equal to two sides and the included angle in the other, each to each: hence they are equal (Prop. V.). But the part cannot be equal to the whole (Ax. 8.); hence, there is no inequality between the sides BA, AC; therefore, the triangle BAC is isosceles.



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

For, make the angle BCD-B. Then, in the
triangle CDB, we shall have CD=BD (Prop. XII.).
Now, the side AC<AD+CD; but AD+CD= C
AD+DB AB: therefore ACAB.

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




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 C<B, it follows, from what has just been proved, that AB<AC; which is contrary to the hypothesis. It the angle CB, then the side AB=AC (Prop. XII.); which is also contrary to the supposition. Therefore, when AB>AC, the angle C must be greater than B.


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

Let A be the point, and DE the given line.


Let us suppose that we can draw two perpendiculars, AB, AC. Produce either of them, as AB, till BF is equal to AB, and Ddraw 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 triangles are equal, and the angle ACB=BCF (Prop. V. Cor.). But the angle ACB is a right angle, by hypothesis; therefore, BCF must likewise be a right angle. But if the adjacent angles BCA, BCF, are together equal to two right angles, ACF must be a straight line (Prop. III.): from whence it follows, that between the same two points, A and F, two straight lines can be drawn, which is impossible (Ax. 11.): hence, two perpendiculars cannot be drawn from the same point to the same straight line.



Scholium. At a given point C, in the line AB, it is equally impossible to erect two perpendiculars to that line. For, if CD, CE, were those two perpendiculars, the angles BCD, BCE, would both be right angles hence they would be equal (Ax. 10.); and the line CD would coincide with CE; otherwise, a part would be equal to the whole, which is impossible (Ax. 8.).





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, drawn on different sides of the perpendicular, cutting off equal distances on the other line, will be equal.

3d, Of two oblique lines, drawn at pleasure, that which is farther from the perpendicular will be the longer.

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


Produce the perpendicular AB till BF is equal to AB, and draw 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 CA are equal (Prop. V. Cor.). But ABF, being a straight line, is shorter than ACF, which is a broken line (Def. 3.); therefore, AB, the half of ABF, is shorter than AC, the half of ACF; hence, the perpendicular is shorter than any oblique line.


Secondly. Let us suppose BC=BE; then will the triangle CAB be equal to the the triangle BAE; for BC=BE, the side AB is common, and the angle CBA-ABE; hence the sides AC and AE are equal (Prop. V. Cor.): therefore, two oblique, lines, equally distant from the perpendicular, are equal.

Thirdly. In the triangle DFA, the sum of the lines AC, CF, is less than the sum of the sides AD, DF (Prop. VIII.); therefore, AC, the half of the line ACF, is shorter than AD, the half of the line ADF: therefore, the oblique line, which is farther from the perpendicular, is longer than the one which is nearer.


Cor. 1. The perpendicular measures the shortest distance of a point from a line.

Cor. 2. From the same point to the same straight line, only two equal straight lines can be drawn; for, if there could be more, we should have at least two equal oblique lines on the same side of the perpendicular, which is impossible.

If from the middle point of a straight line, a perpendicular be drawn to this line;

1st, Every point of the perpendicular will be equally distant from the extremities of the line.

2d, Every point, without the perpendicular, will be unequally distant from those extremities.

Let AB be the given straight line, C the middle point, and ECF the perpendicular.

First, Since AC=CB, the two oblique lines AD, DB, are equally distant from the perpendicular, and therefore equal (Prop. XV.). So, ǎkewise, are the two oblique lines AE, EB, the A two AF, FB, and so on. Therefore every point in the perpendicular is equally distant from the extremities A and B.


Secondly, Let I be a point out of the perpendicular. If IA and IB be drawn, one of these lines will cut the perpendicular in D; from which, drawing DB, we shall have DB=DA. But the straight line IB is less than ID+DB, and ID+DB=ID+DA=IA; therefore, IB<IA; therefore, every point out of the perpendicular, is unequally distant from the extremities A and B.

Cor. If a straight line have two points D and F, equally distant from the extremities A and B, it will be perpendicular to AB at the middle point C.


If two right angled triangles have the hypothenuse and a side of the one, equal to the hypothenuse and a side of the other, each to each, the remaining parts will also be equal, each to each, and the triangles themselves will be equal.


In the two right angled triangles BAC, EDF, let the hypothenuse AC=DF, and the side BA=ED: then will the side BC=EF, the angle A=D, and the angle C-F.



If the side BC is equal to EF, the like angles of the two triangles are equal (Prop. X.). Now, if it be possible, suppose these two sides to be unequal, and that BC is the greater.



On BC take BG=EF, and draw AG. Then, in the two triangles BAG, DEF, the angles B and E are equal, being right angles, the side BA-ED by hypothesis, and the side BG=EF by construction: consequently, AG-DF (Prop. V. Cor.). But, by hypothesis AC-DF; and therefore, AC=AG (Ax. 1.). But the oblique line AC cannot be equal to AG, which lies nearer the perpendicular AB (Prop. XV.); therefore, BC and EF cannot be unequal, and hence the angle A=D, and the angle C=F; and therefore, the triangles are equal (Prop. VI. Sch.).


If two straight lines are perpendicular to a third line, they will be parallel to each other: in other words, they will never meet, how far soever either way, both of them be produced.

Let the two lines AC, BD, A be perpendicular to AB; then will they be parallel.


For, if they could meet in a point O, on either side of AB, there would be two perpendiculars OA, OB, let fall from the same point on the same straight line; which is impossible (Prop. XIV.).



Let the two lines EC, BD, meet the third line BA, making the angles BAC, ABD, together equal to two right angles: then the lines EC, BD, will be parallel.

If two straight lines meet a third line, making the sum of the interior angles on the same side of the line met, equal to two right angles, the two lines will be parallel.



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From G, the middle point of BA, draw the straight line EGF, perpendicular to EC. It will also be perpendicular to BD. For, the sum BAC+ABD is equal to two right angles, by hypothesis; the sum BAC+BAE is likewise equal to two right angles (Prop. I.); and taking away BAC from both, there will remain the angle ABD=BAE.

Again, the angles EGA, BGF, are equal (Prop. IV.); there fore, the triangles EGA and BGF, have each a side and two adjacent angles equal; therefore, they are themselves equal, and the angle GEA is equal to the angle GFB (Prop. VI. Cor.): but GEA is a right angle by construction; therefore, GFB is a right angle; hence the two lines EC, BD, are perpendicular to the same straight line, and are therefore parallel (Prop. XVIII.).

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