Triangles upon

the same base, and between the same parallels, are equal to one

another.

Let the triangles ABC, DBC be upon the same base BC, and between the same parallels, AD, BC: The triangle ABC is equal to the triangle DBC.

Produce AD both ways to the points E, F, and through B draw (31. 1.) BE parallel to CA; and through C draw CF parallel to BD: Therefore, each of the figures EBCA, DBCF is a parallelogram; and EBCA

E

B

A D

F

is equal (35. 1.) to DBCF, because they are upon the same base BC, and between the same parallels BC, EF; but the triangle ABC is the half of the parallelogram EBCA, because the diameter AB bisects (34. 1.) it ; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it; and the halves of equal things are equal (7. Ax.); therefore the triangle ABC is equal to the triangle DBC.

Triangles upon equal bases, and between the same parallels, are equal to one

another.

G

D

H

Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD: The triangle ABC is equal to the triangle DEF. Produce AD both ways to the points G, H, and through B draw BG parallel (31. 1.) to CA, and through F draw FH parallel to ED: Then each of the figures GBCA, DEFH is a parallelogram; and they are equal to (36. 1.) one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half (34. 1.) of the parallelogram GBCA, because

B

CE

F

the diameter AB bisects it; and the triangle DEF is the half (34. 1.) of the parallelogram DEFH, because the diameter DF bisects it: But the halves of equal things are equal (7. Ax.); therefore the triangle ABC is equal to the triangle DEF.

PROP. XXXIX. THEOR.

Equal triangles upon the same base, and upon the same side of it, are between the same parallels.

Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it; they are between the same parallels.

Join AD; AD is parallel to BC; for, if it is not, through the point A

D

E

draw (31. 1.) AE parallel to BC, and join EC: A The triangle ABC, is equal (37. 1.) to the triangle EBC, because it is upon the same base BC, and between the same parallels BC, AE: But the triangle ABC is equal to the triangle BDC; therefore also the triangle BDC 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 may be demonstrated that no other line but AD is parallel to BC; AD is therefore parallel to it.

PROP. XL. THEOR.

B

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

Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and to

wards 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 ABC is equal (38. 1.) to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF,

A

D

G

A

B

C

E

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.

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

Let the parallelogram 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.

Join AC; then the triangle ABC is equal (37. 1.) to the triangle EBC, because they are upon the same base 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

A

B

D

E

C

into two equal parts; wherefore ABCD is also double of the triangle EBC.

PROP. XLII. PROB.

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 have one of its angles equal to D.

the

A

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 BE is equal to EC, 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 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 CEF equal to the given angle D.

BE

D

COR. Hence, if the angle D be a right angle, the parallelogram EFGC will be a rectangle, equivalent to the triangle ABC; and therefore the same construction will apply to the problem: to make a triangle equivalent to a given rectangle.

PROP. XLIII. THEOR.

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

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 complement BK is equal to the complement KD.

A

H

D

E

K

F

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 parallelogram, and AK its diameter, the triangle AEK is equal to the triangle AHK: For the same B reason, the triangle KGC is equal to the

G

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.

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 angle equal to D. Make (42. 1.) the parallelogram BEFG equal to 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 EF, 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 (1 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. 1.) to BF: but BF is equal to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equal (15. 1.) to the angle ABM, and likewise to the angle D; the angle ABM is equal to the angle 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 angle D.

COR. Hence, a triangle may be converted into an equivalent rectangle, having a side of a given length: for, if the angle D be a right angle, and AB the given side, the parallelogram ABML will be a rectangle equiva lent to the triangle C.

PROP. XLV. PROB.

To describe a parallelogram equal to a given rectilineal figure, 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 ADB, 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 because at the point H in the straight lines 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.

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.