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Let ABCDE be any figure; then the sum of all its inward angles, A+B+C+D+ E, is equal to twice as many right angles, wanting four, as the figure has sides.
For, from any point P, within it, draw lines, PA, PB, PC, &c. to all the angles, dividing the polygon into as many triangles as it has sides. Now the sum of the three angles of each of these triangles, is equal to two right angles (th. 17); therefore the sum of the angles of all the triangles is equal to twice as many right angles as the figure has sides. But the sum of all 1 the angles about the point P, which are so many of the angles of the triangles, but no part of the inward angles of the polygon, is equal to four right angles (corol. 3, th. 6), and must be deducted out of the former sum. Hence it follows, that the sum of all the inward angles of the polygon alone, A + B + C +D+ E, is equal to twice as many right angles as the figure has sides, wanting the said four right angles. Q. E. D.
Let A, B, C, &c. be the outward angles of any polygon, made by producing all the sides; then will the sum A+B+C+D+E, of all those outward angles, be equal to four right angles.
When every side of any figure is produced out, the sum of all the outward angles thereby made, is equal to four right angles.
For every one of these outward angles, together with its adjacent inward angle, make up two right angles, as A+ a equal to two right angles, being the two angles made by one line meeting another (th. 6). And there being as many outward, or inward angles, as the figure has sides; therefore the sum of all the inward and outward angles, is equal to twice as many right angles as the figure has sides; therefore the sum of all the inward and outward angles, is equal to twice as many right angles as the figure has sides. But the sum of all the inward angles, with four right angles, is equal to twice as many right angles as the figure has sides (th. 19). Therefore the sum of all the inward and all the outward angles, is equal to the sum of all the inward angles and four right angles (by ax. 1). From each of these take away all the inward angles, and there remain all the outward angles equal to four right angles (by ax. 3). Q. E. D.
If AB, AC, AD, &c. be lines drawn from the given point A, to the indefinite line DE, of which AB is perpendicular; then shall the perpendicular AB be less than AC, and AC less than AD, &c.
For, the angle B being a right one, the angle C is acute, (by cor. 3, th. 17), and therefore less than the angle B. But the less angle of a triangle is subtended by the less side (th. 9). Therefore the side AB is less than the side AC.
A perpendicular is the shortest line that can be drawn from a given point to an indefinite line. And, of any other lines drawn from the same point, those that are nearest the perpendicular are less than those more remote.
Again, the angle ACB being acute, as before, the adjacent angle ACD will be obtuse (by th. 6); consequently the angle D is acute (corol. 3, th. 17), and therefore is less than the angle C. And since the less side is opposite to the less angle, therefore the side AC is less than the side AD. Q. E.D.
Corol. A perpendicular is the least distance of a given point from a line.
The opposite sides and angles of any parallelogram are equal to each other; and the diagonal divides it into two equal triangles.
Let ABCD be a parallelogram, of which the diagonal is BD; then will its opposite sides and angles be equal to each other, and the diagonal BD will divide it into two equal parts, or triangles.
For, since the sides AB and DC are parallel, as also the sides AD and BC (defin. 32), and the line BD meets them; therefore the alternate angles are equal (th. 12), namely,
the angle ABD to the angle CDB, and the angle ADB to the angle CBD. Hence the two triangles, having two angles in the one equal to two angles in the other, have also their third angles equal (cor. 1, th. 17), namely, the angle A equal to the angle C, which are two of the opposite angles of the parallelogram.
Also, if to the equal angles ABD, CDB, be added the equal angles CBD, ADB, the wholes will be equal (ax. 2), namely, the whole angle ABC to the whole ADC, which are the other two opposite angles of the parallelogram. Q. E. D.
Let ABCD be a quadrangle, having the opposite sides equal, namely, the side AB equal to DC, and AD equal to BC; then shall these equal sides be also parallel, and the figure a parallelogram.
Again, since the two triangles are mutually equiangular, and have a side in each equal, viz. the common side BD; therefore the two triangles are identical (th. 2), or equal in all respects, namely, the side AB equal to the opposite side DC, and AD equal to the opposite side BC, and the whole triangle ABD equal to the whole triangle BCD. Q. E. D.
Corol. 1. Hence, if one angle of a parallelogram be a right angle, all the other three will also be right angles, and the parallelogram a rectangle.
Corol. 2. Hence also, the sum of any two adjacent angles of a parallelogram is equal to two right angles.
Every quadrilateral, whose opposite sides are equal, is a parallelogram, or has its opposite sides parallel.
BC, and the figure is a parallelogram. Q. E. D.
For, let the diagonal BD be drawn. Then, the triangles, ABD, CBD, being mutually equilateral (by hyp.), they are also mutually equiangular (th. 5), or have their corresponding angles equal; consequently the opposite sides are parallel (th. 13); viz. the side AB parallel to DC, and AD parallel to
Those lines which join the corresponding extremes of two equal and parallel lines, are themselves equal and parallel.
Let AB, DC, be two equal and parallel lines; then will the lines AD, BC, which join their extremes, be also equal and parallel. [See the fig. above.]
For, draw the diagonal BD. Then, because AB and DC are parallel (by hyp.), the angle ABD is equal to the alternate angle BDC (th. 12). Hence then, the two triangles having two sides and the contained angles equal, viz. the side AB equal to the side DC, and the side BD common, and the contained angle ABD equal to the contained angle BDC, they have the remaining sides and angles also respectively equal (th. 1); consequently AD is equal to BC, and also parallel to it (th. 12). Q. E. D.
fill. 5 X
Parallelograms, as also triangles, standing on the same base, and between the same parallels, are equal to each other.
Let ABCD, ABEF, be two parallelograms, and ABC, ABF, two triangles, standing on the same base, AB, and between the same parallels AB, DE; then will the parallelogram ABCD be equal to the parallelogram ABEF, and the triangle ABC equal to the triangle ABF.
For, since the line DE cuts the two parallels AF, BE, and the two AD, BC, it makes the angle E equal to the angle AFD, and the angle D equal to the angle BCE (th. 14); the two triangles ADF, BCE, are therefore equiangular (cor. 1, th. 17); and having the two corresponding sides AD, BC, equal (th. 22), being opposite sides of a parallelogram, these two triangles are identical, or equal in all respects (th. 2). If each of these equal triangles then be taken from the whole space ABED, there will remain the parallelogram ABEF in the one case, equal to the parallelogram ABCD in the other (by ax. 3).
Also the triangles ABC, ABF, on the same base AB, and between the same parallels, are equal, being the halves of the said equal parallelograms (th. 22). Q. E. D.
Corol. 1. Parallelograms, or triangles, having the same base and altitude, are equal. For the altitude is the same as the perpendicular or distance between the two parallels, which is every where equal, by the definition of parallels.
Corol. 2. Parallelograms, or triangles, having equal bases and altitudes, are equal. For, if the one figure be applied with its base on the other, the bases will coincide or be the same, because they are equal: and so the two figures, having the same base and altitude, are equal.
If a parallelogram and a triangle, stand on the same base, and between the same parallels, the parallelogram will be double the triangle, or the triangle half the parallelogram.
Let ABCD be a parallelogram, and ABE a triangle, on the same base AB, and between the same parallels AB, DE; then will the parallelogram ABCD be double the triangle ABE, or the triangle half the parallelogram.
For, draw the diagonal AC of the parallelogram, dividing it into two equal parts (th. 22). Then because the triangles ABC, ABE, on the same base, and between the same parallels, are equal (th. 25); and because the one triangle ABC is half the parallelogram ABCD (th. 22), the other equal triangle ABE is also equal to half the same parallelogram ABCD. Q. E. D.
Corol. 1. A triangle is equal to half a parallelogram of the same base and altitude, because the altitude is the perpendicular distance between the parallels, which is everywhere equal, by the definition of parallels.
Corol. 2. If the base of a parallelogram be half that of a triangle, of the same altitude, or the base of the triangle be double that of the parallelogram, the two figures will be equal to each other.
Rectangles that are contained by equal lines, are equal to each other.
Let BD, FH, be two rectangles, having the sides AB, BC, equal to the sides EF, FG, each to each; then will the rectangle BD be equal to the rectangle FH.
For, draw the two diagonals AC, EG, dividing the two parallelograms each into two equal parts. Then the two triangles ABC, EFG, are equal to each other (th. 1), because they have the two sides AB, BC, and the contained angle B, equal to the two sides EF, FG, and the contained angle F (by hyp). But these equal triangles are the halves of the respective rectangles. And because the halves, or the triangles, are equal, the wholes, or the rectangles DB, HF, are also equal (by ax. 6). Q. E. D.
Corol. The squares on equal lines are also equal; for every square is a species of rectangle.
B E F
The complements of the parallelograms, which are about the diagonal of any parallelogram, are equal to each other.
Let AC be a parallelogram, BD a diagonal, EIF parallel to AB or DC, and GIH parallel to AD or BC, making AI, IC, complements to the parallelograms EG, HF, which are about the diagonal DB: then will the complement AI be equal to the complement IC.
For, since the diagonal DB bisects the three parallelograms AC, EG, HF (th. 22); therefore, the whole triangle DAB being equal to the whole triangle DCB, and the parts DEI, IHB, respectively equal to the parts DGI, IFB, the remaining parts AI, IC, must also be equal (by ax. 3). Q. E. D.
A trapezoid, or trapezium having two sides parallel, is equal to half a paral
lelogram, whose base is the sum of these two sides, and its altitude the perpendicular distance between them.
Let ABCD be the trapezoid, having its two sides AB, DC, parallel; and in AB produced take BE equal to DC, so that AE may be the sum of the two parallel sides; produce DC also, and let EF, GC, BH, be all three parallel to AD. Then is AF a parallelogram of the same altitude with the trapezoid ABCD, having its base AE equal to the sum of the parallel sides of the trapezoid; and it is to be proved that the trapezoid ABCD is equal to half the parallelogram AF.
Now, since triangles, or parallelograms, of equal bases and altitude, are equal (corol. 2, th. 25), the parallelogram DG is equal to the parallelogram HE, and the triangle CGB equal to the triangle CHB; consequently, the line BC bisects, or equally divides, the parallelogram AF, and ABCD is the half of it. Q.E.D.
The sum of all the rectangles contained under one whole line, and the several parts of another line, any way divided, is equal to the rectangle contained under the two whole lines.
E F B
Let AD be the one line, and AB the other, divided into the parts AE, EF, FB; then will the rectangle contained by AD and AB, be equal to the sum of the rectangles of AD and AE, and AD and EF, and AD and FB: thus expressed, AD. AB = AD. AE + AD. EF + AD. FB. For, make the rectangle AC of the two whole lines AD, AB; and draw EG, FH, perpendicular to AB, or parallel to AD, to which they are equal (th. 22). Then the whole rectangle AC is made up of all the other rectangles AG, EH, FC. But these rectangles are contained by AD and AE, EG and EF, FH and FB; which are equal to the rectangles of AD and AE, AD and EF, AD and FB, because AD is equal to each of the two EG, FH. Therefore, the rectangle AD. AB is equal to the sum of all the other rectangles AD. AE, AD. EF, AD. FB. Q. E. D.
Corol. If a right line be divided into any two parts, the square on the whole line, is equal to both the rectangles of the whole line and each of the parts.
The square of the sum of two lines, is greater than the sum of their squares, by twice the rectangle of the said lines. Or, the square of a whole line is equal to the squares of its two parts, together with twice the rectangle of those parts.
Let the line AB be the sum of any two lines AC, CB; then will the square of AB be equal to the squares of AC, CB, together with twice the rectangle of AC. CB. That is, AB2 = AC2 + CB2 + 2 AC. CB.
For, let ABDE be the square on the sum or whole line AB, and ACFG the square on the part AC. Produce CF and GF to the other sides at H and I.
From the lines CH, GI, which are equal, being each equal to the sides of the