BG parallel to CA (I. 31), and through F draw F H parallel to ED. Then each of the figures GC, EH, is a parallelogram.

Q

B

A

C E

D

I

The parallelograms GC, EH are equal to one another (I. 36), because they are upon equal bases BC, EF, and between the same parallels BF, GH. The triangle ABCis the half of the parallelogram GC (I. 34), because the diagonal AB bisects it. And the triangle DEF is the half of the parallelogram EH, because the diagonal DF bisects it. But the halves of equal things are equal (Ax. 7). Therefore the triangle ABC is equal to the triangle DEF. Wherefore, triangles upon equal bases, &c. Q. E. D.

PROP. XXXIX. (THEOREM.)-Equal triangles (ABC, DBC) upon the same base (BC), and upon the same side of it, are between the same parallels (AD, BC).

For if AD be not parallel to DC, through the point A draw AE parallel to BC (I. 31), meeting BD in E; and join EC.

A.

D

The triangle ABC is equal to the triangle EBC (I. 37) because they are upon the same base BC, and between the same parallels BC, AE. But the triangle ABC is equal (Hyp.) to the triangle DBC. Therefore the triangle DBC is equal to the triangle EBC, the greater to the less, which is impossible. Therefore AE is not parallel to BC. In the same manner it can be proved that no other straight line but AD is parallel to BC. Therefore AD is parallel to BC. Wherefore, equal triangles upon, &c. Q. E. D.

PROP. XL. (PROBLEM.) — Equal triangles (ABC, DEF) upon equal bases (BC, EF) in the same straight line (BF), and on the same side of it, are between the same parallels (AD, BF).

For, if AD be not parallel to BF, through A draw AG parallel to

BF (I. 31), meeting ED in G, and join GF.

A

D

The triangle ABC is equal to the triangle GEF (I. 38), because they are upon equal bases BC, EF, and between the same parallels BF, AG. But the triangle ABC is equal (Hyp.) to the triangle DEF. Therefore the triangle DEF is equal (Ax. 1) to the triangle GEF, the greater equal to the less, which is impossible. Therefore AG is not parallel to BF. In the same manner it can be proved that no other straight line is parallel to BF, but AD. Therefore AD is parallel to BF. Wherefore, equal triangles upon, &c. Q. E. D.

B

PROP. XLI. (THEOREM.)—If a parallelogram (BD) and a triangle (EBC) be upon the same base (BC), and between the same parallels (BC, AE); the parallelogram is double of the triangle.

Join AC. The triangle ABC is equal to the triangle EBC

с

(I. 37), because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram BD is double of the triangle ABC (I. 34), because the diagonal AC bisects it. Therefore the parallelogram BD is also double of the triangle EBC. Therefore, if a parallelogram and a triangle, &c. Q. E. D.

PROP. XLII. (PROBLEM.)-To describe a parallelogram equal to a given triangle (ABC), and having one of its angles equal to a given rectilineal angle (D).

Bisect BC in E (I. 10), and join AE. At the point E in the straight line EC, make the angle CEF (I. 23) equal to the angle D. Through A draw AFG (I. 31), parallel to BC, and through C draw CG parallel to EF. Then the figure CEFG is a parallelogram (Def. 36).

AFG

E

Because the two triangles ABE, AEC are on equal bases BE, EC, and between the same parallels BC, AG; they are equal (1. 38) to one another. Therefore the triangle ABC is B double of the triangle AEC. But the parallelogram FC is double of the triangle AEC (I. 41) because they are upon the same base EC, and between the same parallels EC, AG. Therefore the parallelogram FC is equal (Ax. 6) to the triangle ABC, and it has one of its angles CEF equal to the given angle D. Wherefore, a parallelogram FC has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q. E. D.

PROP. XLIII. (THEOREM.)-The complements (BK, KD) of the parallelograms (EH, GF), which are about the diagonal (AC) of any parallelogram, are equal.

A H

D

F

Because BD is a parallelogram, and AC its diagonal, the triangle ABC is equal (I. 34) to the triangle ADC. Again, because EH is a parallelogram, and AK its diagonal, the triangle AEK is equal (I. 34) to the triangle AHK. For the same reason, the triangle KGC is equal to the triangle KFC. Therefore the two triangles AEK, KGC are equal (Ax. 2) to the two triangles AHK, KFC. But the whole triangle ABC is equal to the whole triangle ADC. Therefore the remaining complement BK is equal (Ax. 3) to the remaining complement KD. Wherefore the complements, &c. Q. E. D.

B G

PROP. XLIV. (PROBLEM.)-To a given straight line (AB) to apply a parallelogram, equal to a given triangle (C), and having one of its angles equal to a given rectilineal angle (D).

Make the parallelogram BF equal (I. 42) to the triangle C, and having the angle EBG equal to the angle D. Place it so that BE

shall be in the same straight line with AB. Produce FG to H. Through A draw AH parallel to BG or EF (I. 31) and meeting FH in H. Join HB.

Because the straight line HF falls upon the two parallels AH, EF, the two angles AHF, HFE are together equal (I. 29) to two right angles. Therefore the two angles BHF, HFE are less than two right angles. But those straight lines which with

another straight line make the

to

M

two interior angles upon the same side of it less than two right angles, meet (4x. 12) if

B

H A

produced far enough. Therefore HB,FE shall meet, if produced. Let them be produced and meet in K. Through K draw KL parallel to EA or FH, and produce HA, GB, to meet KL in the points L and M.

Because FL is a parallelogram (Def. 36), of which the diagonal is HK; AG, ME, are the parallelograms about HK, and LB, BF are the complements. Therefore the complement LB is equal (I. 43) to the complement BF. But the complement BF is equal (Const.) to the triangle C. Therefore LB is equal (Ax. 1) to the triangle C. Because the angle GBE is equal (I. 15) to the angle ABM, and likewise (Const.) to the angle D; therefore the angle ABM is equal (Ax. 1) to the angle D. Wherefore to the straight line AB is applied the parallelogram LB, equal to the triangle C, and having the angle ABM equal to the angle D. Q. E. F.

PROP. XLV. (PROBLEM.)-To describe a parallelogram equal to a given rectilineal figure (ABCD) and having an angle equal to a given rectilineal angle (E).

A

D

F

Join DB. Describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal (I. 42) to the angle E. To the straight line GH apply the parallelogram GM equal to the triangle DBC, and having the angle GHM equal (I. 44) to the angle E. Then the figure KL is the parallelogram required.

Because the angle E is equal (Const.) to each of the angles FKH,

GHM, the angle FKH is equal (Ax. 1) to the angle GHM. To each of these equals add the angle KHG. Therefore the two angles FKH, KHG are equal to the two angles KHG, GHM. But the two angles FKH, KHG are equal (I. 29) to two right angles. Therefore the two angles K HG, GHM, are also equal (Ax. 1) to two right angles. Because at the point H, in the straight line GH, the two straight lines K H, H M, upon opposite sides of it, inake the adjacent angles equal to two right angles, HK is in the same straight line (I. 14) with HM. Again, because the straight line H G meets the parallels KM, FG, therefore the angle MHG is equal (I. 29) to the alternate angle HGF. To each of these equals add the angle HGL.

Therefore the two angles MHG, HGL, are equal to the two angles HGF, HGL; but the two angles MHG, HGL, are equal (I. 29) to two right angles. Therefore also the two angles HGF, HGL, are equal (4x. 1) to two right angles. Wherefore, as before, FG (I. 14) is in the same straight line with GL. Because KF is parallel to HG, and HG to ML, KF is parallel (I. 30) to ML. And KM has been proved parallel to FL. Therefore the figure KL is a parallelogram (Def. 36). But the parallelogram FH is equal (Const.) to the triangle ABD, and the parallelogram GM to the triangle BDC. Therefore the whole parallelogram KL is equal to the whole rectilineal figure ABCD. Wherefore the parallelogram KL has been described equal to the given rectilineal figure ABCD, and having the angle F KM equal to the given angle E. Q. E. F.

PROP. XLVI. (PROBLEM.)--To describe a square upon a given straight line (A B).

From the point A draw A C at right angles (I. 11) to AB Make AD equal (I. 3) to A B. Through the point D draw DE parallel (I. 31) to AB, and through the point B draw BE parallel to AD. The figure AE is a square.

୯

A

B

E

Because A E is a parallelogram (Def. 36) AB is equal (I. 34) to DE, and AD to BE. But BA is equal to AD. Therefore the four lines BA, AD, D DE, EB, are all equal to one another, and the parallelogram AE is equilateral. Again, because AD meets the parallels AB, D E, the two angles BAD, ADE, are equal (I. 29) to two right angles. But BAD is (Const.) a right angle. Therefore, also, ADE is a right angle. But the opposite angles of parallelograms (I. 34) are equal. Therefore each of the opposite angles ABE, BED, is a right angle, and the parallelogram AE is rectangular, that is, has all its angles right angles. And it has been proved to be equilateral. Therefore the figure AE is a square (Def. 30), and it is described upon the given straight line AB. Q.E. F.

Corollary 1.-Hence, every parallelogram that has one right angle has all its right angles.

PROP. XLVII. (THEOREM.)—In any right-angled triangle (ABC), the square described upon the side (B C) subtending the right angle (BAC) is equal to the squares described upon the two sides (B A and À C) containing the right angle.

Upon the side BC describe the square BE (I. 46), and on the sides BA and AC, the squares GB and HC. Through A draw AL parallel to BD or CE (I. 31), and join AD and FC.

Because the angle BAC is a right angle (Hyp), and that the angle BAG is a right angle (Def. 30), the two straight lines AC, AG, upon the opposite sides of AB, make with it, at the point A, the adjacent angles equal to two right argles. Therefore CA is (I. 14) in the same straight line with AG. For the same reason, BA and AH are in the same straight line. The angle DBC is equal to the angle FBA, each of

H

them being a right angle. To each of these equals add the angle ABC Therefore the whole angle DBA is equal (Ax. 2) to the whole angle FBC. And because the two sides AB, BD, are equal to the two sides FB, BC, each to each, and the angle ABD to the anglo FBC; therefore the base AD is equal to the base FC (I. 4), and the triangle ABD to the triangle FBC. But the parallelogram BL is double of the triangle ABD (I. 41), because they are upon the same base BD, and between the same parallels BD and AL. Also the square GB is double of the triangle FBC, because these are upon the same base FB, and between the same parallels FB and GC. Now the doubles of equals are equal to one another (4x. 6). Therefore the parallelogram BL is equal to the square GB. Similarly, by joining

B

AE and BK, it can be proved that the parallelogram CL is equal to the square HC. Therefore the whole square BE is equal (Ax. 2) to the two squares GB and HC. Wherefore the square described upon the side BC, opposite to the right to the squares described upon the two sides AB it. Therefore, in any right-angled triangle, &c.

angle BAC, is equal and AC, containing Q. E. D.

PROP. XLVIII. (THEOREM.)—If the square described upon one of the sides (BC) of a triangle (ABC), be equal to the squares described upon the other two (AB and AC) sides of it; the angle (BAC) contained by these two sides is a right angle.

From the point A draw AD at right angles (I. 11) to AC. Make AD equal (I. 3) to AB, and join DC.

A

D

C

Because AD is equal to AB, the square of AD is equal to the square of AB. To each of these equals add the square of AC. Therefore the squares of AD and AC are equal (Ax. 2) to the squares of AB and AC. But the squares of AD and AC are equal to the square of DC (I. 47), because the angle DAC is a right angle. And the square of BC (Hyp) is equal to the squares of BA and AC. Therefore the square of DC is equal (Ax. 1) to the square of BC, and the side DC to the side BC. Again, because the side AD is equal to the side AB, and AC common to the two triangles DAC, BAC, the two sides DA, AC are equal to the two sides BA, AC, each to each, but the base DC has been proved to be equal to the base BC. Therefore the angle DAC is equal (I. 8) to the angle BAC. But DAC is a right angle. Therefore also BAC is a right angle. Therefore, if the square described upon, &c. Q. E. D.