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PROPOSITION XLI. THEOREM.

If a parallelogram and a triangle be on the same base, and between the same parallels, the parallelogram is double of the triangle.

Let the paral. ABCD and the A triangle EBC be on the same base BC, and between the same parallels BC, AE; the paral. ABCD is double of the triangle EBC.

D E

Join AC; then the triangle ABC is equal to EBC (Prop. 37). But the paral. ABCD is double of the triangle ABC (Prop. 34); wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelogram &c. Q. E D.

B

C

COR. If a parallelogram and a triangle be between the same parallels, and if the base of the paral. be half the base of the triangle, the paral. will be equal to the triangle.

For if a paral. were described on half the base BC, and between the parallels BC, AE, it is evident that it would be equal to half the paral. ABCD, and therefore equal to the triangle EBC on double the base.

PROPOSITION XLIII.

THEOREM.

ED.

The complements of the two parallelograms which are about the diagonal of any parallelogram are equal to each other.

A H

D

Let ABCD be a paral. of which the diagonal is AC; let EH, FG be the parals. about AC, tha is, through which AC passes; and let BK, KD be the other parals. which complete the whole figure ABCD, and are therefore called the comple- E ments. The complements BK, KD are equal.

B G

K

Because ABCD, AEKH, KGCF are parallelograms, and AC is their common diagonal, the triangle ABC is equal to ADC (Prop. 34), and the triangle AEK to AHK, and the triangle KGC to KFC; therefore the complements BK, KD are equal (3 Ax.). Therefore, the complements &c. Q. E. D.

H

ED.

PROPOSITION XLVI. PROBLEM.

To describe a square on a given straight line.

Let AB be the given straight line on which it is required to describe a square.

From the point A draw AC perp. to AB, and make AD equal to AB; through D draw DE parallel to AB (Prop. 31), and through B draw BE parallel to AD; then ADEB is a square described on AB. For ADEB is a paral., therefore AB is equal to DE (Prop. 34), and AD to BE. But BA is equal to AD; therefore the four lines BA, AD, DE, EB, are equal to one ano- C ther, and the paral. ADEB is equila

teral.

The line AD meeting the parallels D AB, DE, makes the angles BAD, ADE equal to two right angles (Prop. 29). But BAD is a right angle; therefore ADE is a right angle. Now the angle B is equal to its opposite angle D, and the angle E to A (Prop. 34); therefore each of the angles B, E is a right angle; where

A

E

B

fore the figure ADEB is rectangular. But it is equilateral; therefore it is a square; and it is described on the given straight line AB. Which was to be done.

COR. A parallelogram having one angle right, and two adjacent sides equal, is a square.

PROPOSITION XLVII. THEOREM.

See Prop. B, Book II.

PROPOSITION C. THEOREM.

ED.

If straight lines be drawn from the same point to meet a given straight line, the perpendicular is the least, and any line which is nearer the perpendicular than another is less than the other,

Let A be any point, and EF a given straight line; and let AB, AC, AD be straight lines drawn from A to meet EF.

Let AB be perp. to EF; AB is less than ‍AC, and AC is less than AD. For since B is a right angle, ACB is an acute angle (3 Cor. 32), Etherefore the side AC is greater than AB (Prop. 19).

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Again, because ACB is an acute angle, ACD is obtuse (Prop. 13), therefore the angle ADC is acute, therefore the side AD is greater than AC, and still greater than AB. Therefore, if straight lines &c. Q. E. D.

Cor. The distance between a point and a straight line is the perpendicular drawn from the point to the line.

Scholium. The term distance signifies the shortest interval between two points; therefore the distance between two points is the straight line which joins them.

PROPOSITION D. THEOREM.

ED.

If the sides of one angle be parallel to the sides of another angle, each to each, and be directed the same way, the angles are equal to each other.

Let the sides AC, AB of the angle BAC be parallel to the sides ED, EF of the angle DEF, each to each; the angle BAC is equal to the angle DEF.

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Produce FE to meet AC in G (in the first figure), then, because BA is parallel to FG, the angle BAC is equal to FGC; and because AC is parallel to ED, the angle FED is equal to FGC; therefore the angle BAC is equal to FED.

Again (in the second figure), because AC is parallel to ED, the angle BAC is equal to the alternate angle AGE (Prop. 29); and because AB is parallel to EF, the angle FED is equal to AGE; therefore the angle BAC is equal to FED. Therefore, if the sides &c. Q. E. D.

PROPOSITION E. THEOREM.

ED.

If two parallelograms have one angle in one parallelogram equal to one angle in the other, the remaining angles of the former are equal to the remaining corresponding angles of the latter.

Let ABCD, EFGH be two parallelograms, which have the angles at A and E equal; the other angles of the former paral. are equal to the other angles of the latter, each to each, namely, the angle B is equal to F, the angle C to G, and the angle D to H.

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For in the paral. ABCD the two angles A and B are together equal to two right angles (Prop. 29); and in the paral. EFGH the two angles E and F are together equal to two right angles. Therefore the angles A and B are together equal to the angles E and F. Therefore if from these equals the equal angles A and E be taken, the remaining angles B and F will be equal.

Again, in the paral. ABCD the opposite angles A and C are equal (Prop. 34), and in the paral. EFGH the opposite angles E and G are equal. But the angles A and E are equal, therefore the angles C and G are equal. In the same manner it may be proved that the angle D, to which B is opposite, is equal to the angle H, to which F is opposite. Therefore, if two parallelograms &c. Q. E. D.

PROPOSITION F. THEOREM.

ED.

If two parallelograms have two adjacent sides of one equal to two adjacent sides of the other, each to each; and have also the angles contained by those sides equal to each other; the parallelograms will be equal in all respects.

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Let ABCD, EFGH be two parallelograms, which have the two sides AB, AD equal to the two sides EF, EH, each to each, and the angle A equal to E; the parallelograms are equal, and the other corresponding sides are equal, and the other corresponding angles are equal.

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Draw the diagonals BD, FH, then the triangle ABD is equal to BCD (Prop. 34), and the triangle EFH is equal to FGH. But the triangle ABD is equal to EFH (Prop. 4), therefore the triangle BCD is equal to FGH. Consequently the paral. ABCD is equal to EFGH.

Again, the side AB is equal to DC, and the side AD is equal to BC (Prop. 34); also the side EF is equal to HG, and the side EH is equal to FG. But the side AB is equal to EF, and the side AD is equal to EH. Consequently the side DC is equal to HG, and the side BC is equal to FG.

Lastly, the angle A is equal to C, and the angle E is equal to G. But the angle A is equal to E, therefore the angle C is equal to G. Consequently the two angles ABC, ADC are together equal to the two angles EFG, EHG together (6 Cor. 32). Hence those four angles are equal to one another. Therefore, if two parallelograms &c. Q. E.D.

COR. 1. Two rectangles contained by equal straight lines are equal to each other.

COR. 2. Squares which stand on equal straight lines are equal to one another.

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