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the side EF. Also, because AC is equal to DF, and BC equal to EF (hyp.), the point A will coincide with the point D, and the point B with the point E; consequently the side AB will coincide with the side DE. Therefore the two triangles are identical, and have all their other corresponding parts equal (ax. 9), namely, the side AB equal to the side DE, the angle A to the angle D, and the angle B to the angle E.
When two triangles have two angles and the included side in the one, equal to two angles and the included side in the other, the triangles are identical, or have their other sides and angle equal.
LET the two triangles ABC, DEF, have the angle A equal to the angle D, the angle B equal to the angle E, and the side AB equal to the side DE; then these two triangles will be identical.
For, conceive the triangle ABC to be placed on the triangle DEF, in such manner that the side AB may fall exactly on the equal side DE. Then, since the angle A is equal to the angle D (hyp.), the side AC must fall on the side DF; and, in like manner, because the angle B is equal to the angle E, the side BC must fall on the side EF. Thus the three sides of the triangle ABC will be exactly placed on the three sides of the triangle DEF; consequently the two triangles are identical (ax 9), having the other two sides AC, BC, equal to the two DF, EF, and the remaining angle C equal to the remaining angle F.
In an isosceles triangle, the angles at the base are equal: or, if a triangle have two sides equal, their opposite angles will also be equal.
If the triangle ABC have the side AC equal to the side
BC: then will the angle B be equal to the angle A.
For, conceive the angle C to be bisected, or divided into two equal parts, by the line CD, making the angle ACD equal to the angle BCD.
Then, the two triangles, ACD, BCD, have two sides and the contained angle of the one, equal to two sides and the contained angle of the other, viz. the side AC equal to BC, the angle ACD equal to BCD, and the side CD common; therefore these two triangles are identical, or equal in all respects (th. 1); and consequently the angle A equal to the angle B.
Cor. 1. Hence the line which bisects the vertical angle of an isosceles triangle, bisects the base, and is also perpendicular to it.
Cor. 2. Hence too it appears, that every equilateral triangle, is also equiangular, or has all its angles equal.
When one line meets another, the angles which it makes on the same side of the other, are together equal to two right angles.
LET the line AB meet the line CD: then will the two angles ABC, ABD, taken together, be equal to two right angles.
For, first, when the two angles ABC, ABD, are equal to each other, they are both of them right angles (def. 25).
But when the angles are unequal, suppose BE drawn perpendicular to CD. Then, since the two angles EBC, EBD, are right angles (def. 25), and the angle EBD is equal to the two angles EBA, ABD, together (ax. 8), the three angles, EBC, EBA, and ABD, are equal to two right angles.
But the two angles EBC, EBA, are together equal to the angle ABC (ax. 8). Consequently the two angles ABC, ABD, are also equal to two right angles.
Cor. 1. Hence also, conversely, if the two angles ABC, ABD, on both sides of the line AB, make up together two right angles, then CB and BD form one continued right line CD.
Cor. 2. Hence, all the angles which can be made, at any point B, by any number of lines, on the same side of the right line CD, are, when taken all together, equal to two right angles.
Cor. 3. And, as all the angles that can be made on the other side of the line CD are also equal to two right angles; therefore all the angles that can be made quite round a point B, by any number of lines, are equal to four right angles. Cor. 4. Hence also the whole circumference of a circle,
being the sum of the measures of all the angles that can be made about the centre F (def. 57), is the measure of four right angles. Consequently, a semicircle, or 180 degrees, is the measure of two right angles; and a quadrant, or 90 degrees, the measure of one right angle.
When two lines intersect each other, the opposite angles are equal.
LET the two lines AB, CD, intersect in the point E; then will the angle AEC be equal to the angle BED, and the angle AED equal to the angle CEB.
For, since the line CE meets the line AB, the two angles AEC, BEC, taken together, are equal to the two right angles (th. 4).
In like manner, the line BE, meetiug the line CD, makes the two angles BEC, BED, equal to two right angles.
Therefore the sum of the two angles AEC, BEC, is equal to the sum of the two BEC, BED (ax. 1).
And if the angle BEC, which is common, be taken away from both these, the remaining angle AEC will be equal to the remaining angle BED (ax. 3).
And in like manner it may be shown, that the angle AED is equal to the opposite angle BEC.
When one side of a triangle is produced, the outward angle is greater than either of the two inward opposite angles.
LET ABC be a triangle, having the side AB produced to D; then will the outward angle CBD be greater than either of the inward opposite angles A or C.
For, conceive the side BC to be bisected in the point E, and draw the line AE, producing it till EF be equal to AE; and join BF.
Then, since the two triangles AEC, BEF, have the side AE the side EF, and the side CE the side BD (suppos.), and the included or opposite angles at E also equal (th. 5), therefore those two triangles are equal in all respects (th. 1), and have the angle C = the corresponding angle EBF. But the angle CBD is greater than the angle EBF; consequently, the said outward angle CBD is also greater than the angle C.
In like manner, if CB be produced to G, and AB be bisected, it may be shown that the outward angle ABG, or its equal CBD, is greater than the other angle A.
When a triangle has two of its angles equal, the sides opposite to them are also
FOR draw CD perpendicular to the base; and conceive the triangle BCD to revolve about CD till the plane of it coincides with that of CDA. Then since CDA and CDB are right angles (def. 25), the line BD will coincide with DA. Now if B coincide with A, the line CB will coincide with CA, and the angles CBD, CAD coinciding, they will be
equal: but if it be denied that B will coincide with A, let it coincide with some other point E in the line DA, and join CE, and hence (th. 6) the angle CED (that is CBD) is greater than CAD, which is contrary to the hypothesis, and hence cannot be true of the figure respecting waich the hypothesis is made. Hence B cannot but coincide with A, and CA cannot but be equal to CB. Cor. Hence every equiangular triangle is also equilateral.
When two triangles have all the three sides in the one, equal to all the three sides in the other, the triangles are identical, or have also their three angles equal, each to each.
LET the two triangles ABC, ABD, have their three sides respectively equal, viz. the side AB equal to AB, AC to AD, and BC to BD; then shall the two triangles be identical, or have their angles equal, viz. those angles that are opposite to the equal sides; namely, the angle BAC to the angle BAD, the angle ABC to the angle ABD, and the angle C to the angle D.
For, conceive the two triangles to be joined together by their longest equal sides, and draw the line CD.
Then, in the triangle ACD, because the side AC is equal to AD (hyp.), the angle ACD is equal to the angle ADC (th. 3). In like manner, in the triangle BCD, the angle BCD is equal to the angle BDC, because the side BC is equal to BD. Hence then, the angle ACD being equal to the angle ADC, and the angle BCD to the angle BDC, by equal additions the sum of the two angles ACD, BCD, is equal to the sum of the two ADC, BDC (ax. 2), that is the whole angle ACB equal to the whole angle ADB.
Since then, the two sides AC, CB, are equal to the two sides AD, DB, each to each (hyp.), and their contained angles ACB, ADB, also equal, the two triangles ABC, ABD, are identical (th. 1), and have the other angles equal, viz. the angle BAC to the angle BAD, and the angle ABC to the angle ABD.
The greater side of every triangle is opposite to the greater angle; and the greater angle opposite to the greater side.
LET ABC be a triangle, having the side AB greater than the side AC; then will the angle ACB, opposite the greater side AB, be greater than the angle B, opposite the less side AC.
For, on the greater side AB, take the part AD equal to
the less side AC, and join CD. Then, since BCD is a triangle, the outward angle ADC is greater than the inward opposite angle B (th. 6). But the angle ACD is equal to the said outward angle ADC, because AD is equal to AC (th. 3). Consequently, the angle ACD also is greater than the angle B. And since the angle ACD is only a part of ACB, much more must the whole angle ACB be greater than the angle B.
Again, conversely, if the angle C be greater than the angle B, then will the side AB, opposite the former, be greater than the side AC, opposite the latter.
For, if AB be not greater than AC, it must be either equal to it, or less than it. But it cannot be equal, for then the angle C would be equal to the angle B (th. 3), which it is not, by the supposition. Neither can it be less, for then the angle C would be less than the angle B, by the former part of this; which is also contrary to the supposition. The side AB, then, being neither equal to AC, nor less than it, must necessarily be greater.
The sum of any two sides of a triangle is greater than the third side.
LET ABC be a triangle; then will the sum of any two of its sides be greater than the third side, as for instance, ACCB greater than AB.
For, produce AC till CD be equal to CB, or AD equal to the sum of the two AC + CB; and join BD:—then, because CD is equal to CB (constr.), the angle D is equal to the angle CBD (th. 3). But the angle ABD is greater than the angle CBD, consequently it must also be greater than the angle D. And, since the greater side of any triangle is opposite to the greater angle (th. 9), the side AD (of the triangle ABD) is greater than the side AB. But AD is equal to AC and CD, or AC and CB, taken together (constr.); therefore AC+ CB is also greater than AB.
Cor. The shortest distance between two points is a single right line drawn from the one point to the other.
The difference of any two sides of a triangle is less than the third side.
LET ABC be a triangle; then will the difference of any two sides, as AB
AC, be less than the third side BC. For, produce the less side AC to D, till AD be equal to the greater side AB, so that CD may be the difference of the two sides AB - AC; and join BD. Then, because AD is equal to AB (constr.) the opposite angles D and ABD are
equal (th. 3). But the angle CBD is less than the angle ABD, and consequently also less than the equal angle D. And since the greater side of any triangle is opposite to the greater angle (th. 9), the side CD (of the triangle BCD) is less than the side BC.
Otherwise. Set off upon AB, a distance AI, equal to AC. Then (th. 20) ACCB is greater than AB, that is, greater than AI + IC. From these, take away the equal parts AC, AI, respectively; and there remains CB greater than IC. Consequently, IC is less than CB.
When a line intersects two parallel lines, it makes the alternate angles equal to each other.
LET the line EF cut the two parallel lines AB, CD; then will the angle AEF be equal to the alternate angle EFD.
For if they are not equal, one of them must be greater than the other; let it be EFD for instance, which is the greater, if possible; and conceive the line FB to be drawn, cutting off the part or angle EFB equal to the angle AEF, and meeting the line AB in the point B.
Then, since the outward angle AEF, of the triangle BEF, is greater than the inward opposite angle EFB (th. 6); and since these two angles also are equal (constr.), it follows, that those angles are both equal and unequal at the same time which is impossible. Therefore, the angle EFD is not unequal to the alternate angle AEF, that is, they are equal to each other.
Cor. Right lines which are perpendicular to one of two parallel lines, are also perpendicular to the other.
When a line, cutting two other lines, makes the alternate angles equal to
LET the line EF, cutting the two lines AB, CD, make the alternate angles AEF, DFE, equal to each other; then will AB be parallel to CB.
For if they be not parallel, let some other line, as FG, be parallel to AB. Then, because of these parallels, the angle AEF is equal to the alternate angle EFG (th. 12). But the angle AEF is equal to the angle EFD (hyp.) Therefore the angle EFD is equal to the angle EFG (ax. 1); that is, a part is equal to the whole: which is impossible. Therefore no line but CD can be parallel to AB.
Cor. Those lines which are perpendicular to the same line, are parallel to each other.