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PROP. XL. THEOR.

Equal triangles on the same side of bases, which are equal and in the same straight line, are between the same parallels.

A

Let the equal triangles ABC, DEF be upon equal basis BC, EF, in the same straight line BF, and towards the same parts; they are between the same parallels.

Join AD; AD is parallel to BC: For, if it is not, through A draw (31. 1.) AG parallel to BF, and join GF. The triangle B

D

G

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ABC is equal (38. 1.) to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG: But the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible: Therefore AG is not parallel to BF; and in the same manner it may be demonstrated that there is no other parallel to it but AD; AD is therefore parallel to BF. Wherefore equal triangles, &c. Q. E. D.

+ PROP. XLI. THEOR.

If a parallelogram and a triangle be upon the same base, and between the same parallel; the parallelogram is double of the triangle.

A

Let the paralellogram ABCD and the triangle EBC be upon the same base BC and between the same parallels BC, AE; the parallelogram ABCD is double of the triangle EBC.

D

Join AC; then the triangle ABC is equal (37. 1.) to the triangle EBC, because they are upon the same hase BC, and between the same parallels BC, AE. But the parallelogram ABCD is double (34. 1.) of the triangle ABC, because the diameter AC divides it into two equal parts; wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelogram, &c. Q. E. D. PROP. XLII. PROB.

B

E

To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let ABC be the given triangle, and D the given rectilineal angle,

It is required to describe a parallelogram that shall be equal to the given triangle ABC, and bave one of its angles equal to D.

F

G

Bisect (10. 1.) BC in E, join AE, and at the point E in the straight line EC make (23. 1.) the angle CEF equal to D; and through A draw (31. 1.) AG parallel to BC, and through C draw CG (31.1.) parallel to EF: Therefore FECG is a parallelogram: And because A BE is equal to EC: the triangle ABE is likewise equal (38. 1) to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC AG; therefore the triangle ABC is double of the triangle AEC. And the parallelogram FECG is likewise double (41. 1.) of the

B

E

triangle AEC, because it is upon the same base, and between the same parallels: Therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D: Wherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of its angles ĈEF equal to the given angle D. Which was to be done.

PROP. XLIII. THEOR.

The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one

another.

A

Let ABCD be a parallelogram of which the diameter is AC; let EH, FG be the parallelograms about AC, that is, through which AC passes, and let BK, KD be the other parallelograms, which make up the whole figure ABCD and are therefore called the complements: The E complement BK is equal to the complement KD.

Because ABCD is a parallelogram and AC its diameter, the triangle ABC is equal (34. 1.) to the triangle ADC: And because EKHA is a pa- B rallelogram and AK its diameter, the

G

H

K

triangle AEK is equal to the triangle AHK: For the same reason, the triangle KGC is equal to the triangle KFC. Then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to the triangle KFC; the triangle AEK, together with the triangle KGC is equal to the triangle AHK, together with the triangle KFC : But the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK is equal to the remaining complement KD. Wherefore, the complements, &c. Q. E. D.

PROP. XLIV. PROB.

To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an an gle equal to D. Make (42. 1.) the parallelogram BEFG equal to

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the triangle C, having the angle EBG equal to the angle D, and the side BE in the same straight line with AB: produce FG to H, and through A draw (31. 1.) AH parallel to BG or F, and join HB. Then because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE, are together equal (29. 1.) to two right angles; wherefore the angles BHF, HFE are less than two right angles: But straight lines which with another straight line make the interior angles, upon the same side, less than two right angles, do meet if produced (Cor. 29. 1.): Therefore HB, FE will meet, if produced; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB to the points L, M: Then HLKF is a parallelogram, of which the diameter is HK, and AG, ME are the parallelograms about HK; and LB, BF are the complements; therefore LB is equal (43. 4.) to BF: but BF is equal to the triangle C; wherefore LB' is equal to the triangle C; and be cause the angle GBE is equal (15. 1.) to the angle ABM, and likewise to the angle D; the angle ABM is equal to the augle D: Therefore the parallelogram LB, which is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the an gle D: Which was to be done.

PROP. XLV. PROB.

To describe a parallelogram equal to a given rectilineal fi gure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.

Join DB, and describe (42. 1.) the parallelogram FH equal to the triangle AĹB, and having the angle HKF equal to the angle E; and

to the straight line GH (44. 1.) apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. And because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM; add to each of these the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal (29. 1.) to two right angles; therefore also KHG, GHM are equal to two right angles: and

D

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because at the point H in the straight line GH, the two straight lines KH, HM, upon the opposite sides of GH, make the adjacent angles equal to two right angles, KH is in the same straight line (14. 1.) with HM. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal (29. 1.); add to each of these the angle HGL; therefore the angles MHG, HGL, are equal to the angles HGF, HGL: But the angles MHG, HGL, are equal (29. 1.) to two right angles; wherefore also the angles HGF, HGL, are equal to two right angles, and FG is therefore in the same straight line with GL. And because KF is parallel to HG, and HG to ML, KF is parallel (30. 1.) to ML; but KM, FL are parallels: wherefore KFLM is a parallelogram. And because the triangle ABD is equal to the parallelogram HF and the triangle DBC to the parallelogram GM, the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

a

COR. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying (44. 1.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. + PROP. XLVI. PROB.

To describe a square upon a given straight line. Let AB be the given straight line: it is required to describe a square upon AB.

From the point A. draw (11. 1.) AC at right angles to AB; and make (3. 1.) AD equal to AB, and through the point D draw DE parallel (31. 1.) to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram: Whence AB is equal (34. 1.) to DE

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and AD to BE: but BA is equal to AD; C
therefore the four straight lines BA, AD,
DE, EB are equal to one another, and
th parallelogram ADEB is equilateral;,
it is likewise rectangular; for the straight D
line AD meeting the parallels, AB. DE,
makes the angles BAD, ADE equal (29.
1.) to two right angles; but BAD is a right
angle; therefore also ADE is a right angle
now the opposite angles of parallelograms
are equal (34. 1.); therefore each of the A
opposite angles ABE, BED is a right an-
gle; wherefore the figure ADEB is rect-

E

B

angular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB: Which was to be done.

COR, Hence every parallelogram that has one right angle has all its angles right angles.

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PROP. XLVII. THEOR.

In any right angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Let ABC be a right angled triangle having the right angle BAC; the square described upon the side BC is equal to the squares described upon BA, AC.

G

H

On BC describe (46. 1.) the square BDEC, and on BA, AC the squares GB, HC; and through A draw (31. 1.) AL parallel to BD or CE and join AD, FC; then, because each of the angles BAC, BAG is a right angle (25. def.), 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 angles; therefore CA is in the same straight line 14. 1.) with AG; for the same reason, AB and AH are in the same straight line, Now because the angle DBC is equal to the angle FBA, each of them being a right angle, adding to each the angle ABC, the whole augie DBA will be equal (2. Ax.) to the whole FBC; and because the two sides AB. BD, are equal to the two FB, BC each, to each and the angle DBA equal to the

B

D

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E

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