BD are equal (3), and also the angles CAD and BDA (3) Prop. 4. (3); therefore the right line AD, cutting the right lines AC and BD makes the alternate angles equal, (4) prop. 27. and therefore the right lines AC and BD are parallel

(4).

PROP. XXXIV. THEOR.

See N.

The opposite sides (AB and CD, AC and BD) of Fig 51. a parallelogram (AD), are equal to one another, as are also the opposite angles (A and D, C and B), and the parallelogram itself is bisected by its diagonal (AD).

For in the triangles CDA, BAD, the alternate angles CDA and BAD, CAD and BDA are equal to one another (1), and the side AD between the equal (1) Prop. 20. angles is common to both triangles, therefore the sides

CD and CA are equal to AB and BD (2), and the (2) Prop. 26. triangle CDA is equal to the triangle BAD (3), and (3) Schol. the angles ACD and ABD are also equal (2); and Prop. 26. since the angle ACD with CAB is equal to two right

angles (4), and ABD with CDB is equal to two right (4) prop. 29. angles, take away from both the equals ACD and ABD, and there remains the angle CAB equal to CDB.

Cor. 1. If one angle of a parallelogram be a right angle, the other three are right angles. For the adjacent angle is right, because with a right angle it is

equal to two right angles (1), and the opposite are (1) Prop. 29. right angles, because they are equal to these right (2) Prop. 34. angles (2).

Cor. 2. If two parallelograms have one angle equal

in each, the other angles are respectively equal. For the angles opposite these equal angles are equal to

them (1), and therefore to each other (2), but the (1) Prop. 34. angles adjacent to them are also equal to each other, (2) Ar. 1. as being together with these equals equal to two (3) Prop. 28. right angles (3).

PROP. XXXV. THEOR.

Parallelograms (BD and BF) on the same base, and Fig. 52. 53. between the same parallels, are equal.

Produce the side BC to G.

and 54.

See N.

Because the lines AB and DC are parallel, the (1) Prop. 29. angles ABG and DCG are equal (1), and because EB and FC are parallel, the angles EBG and FCG are also equal (1), therefore the angles ABE and (2) Ax. 3. DCF are equal (2), and the sides AB and BE are (3) Prop. 34. equal to DC and CF (3), therefore the triangles A BE (4) Prop.4. and DCF are equal (4); take away the triangle A BE from the quadrilateral ABCF, the remainder is the parallelogram BF, and from the same quadrilateral take away the triangle DCF, the remainder is the parallelogram BD, therefore the parallelograms BF and BD are equal (5).

(5) Ax. 3.

Fig. 55.

Prop.34.

PROP. XXXVI. THEOR.

Parallelograms (BD and EG) on equal bases, and between the same parallels, are equal.

Draw the right lines BF and CG.

Because the lines BC and FG are equal to the same (1) hypoth. & EH (1), they are equal to one another, but they are also parallel (2), therefore BF and CG which join (2) Hypoth. (3) Prop. 33. their extremities are parallel (3), and BG is a paral(4) prop-35. lelogram, therefore equal to both BD and EG (4), and therefore the parallelograms BD and EG are (5) Ar. 1. equal (5).

Fig. 56.

PROP. XXXVII. THEOR.

Triangles (BAC and BDC) on the same base and between the same parallels are equal.

Draw through the point C the right lines CE and (1) Prop. 31. CF parallel to BA and BD (1), the parallelograms (2) Prop. 35. BAEC, BDFC are equal (2), but the triangles BAC (3) Prop. 34. and BDC are halves of them (3), and therefore equal (4) Ax. 7.

(4).

PROP. XXXVIII. THEOR.

Triangles (BAC and HDE) on equal bases and between Fig. 57. the same parallels, are equal.

Draw through the points C and E, the right lines CF (1) Prop.31 and EG parallel to BA and HD(1); the parallelograms (2) Prop.36 BAFC, HDGE are equal (2), but the triangles BAC (3) Prop.34 and HDE are halves of them(3), and therefore equal(4). (4) Ax. 7.

PROP. XXXIX. THEOR.

Equal triangles (BAC and BDC) on the same base Fig.58. and on the same side of it are between the same parallels.

For if the right line AD which joins the vertices of the triangles be not parallel to BC, draw through the point A a right line AF parallel to BC, cutting a side BD of the triangle BDC, or the side produced, in a point E different from the vertex, and draw CE.

Because the right lines AF and BC are parallel, the triangle BEC is equal to BAC (1), but BDC is also (1) Prop.37 equal to BAC (2), therefore BEC and BDC are equal (2) hypoth. (3) a part equal to the whole, which is absurd. There- (3) 4r. 1. fore the line AF is not parallel to BC, and in the same mannerit can be demonstrated that no other line except AD is parallel to it, therefore AD is parallel to BC.

PROP. XL. THEOR.

Equal triangles (BAC and GDH) on equal bases and Fig. 59. on the same side are between the same parallels.

For if the right line AD which joins the vertices of the two triangles be not parallel to BH, draw through the point A the right line AF parallel to BH, cutting

E

See N.

a side GD of the triangle GDH or the side produced, in a point E different from the vertex, and join HE.

Because the right line AF is parallel to BH, and BC (1) Prop.38 and GH are equal, the triangleGEH is equal to BAC(1), (2) hypoth. but GDH is also equal to BAC (2), therefore GEH (3) Ax. 1. and GDH are equal (3), a part equal to the whole,

Fig. 60.

which is absurd. Therefore AF is not parallel to BH, and in the same manner it can be demonstrated, that no other line, except AD, is parallel to BH, therefore AD is parallel to BH.

PROP. XLI. THEOR.

If a parallelogram (BF) and a triangle (BAC) have the same base and be between the same parallels, the parallelogram is double of the triangle.

Draw CD. The triangle BDC is equal to the trian(1) Prop.37 gle BAC(1), but BF is double of the triangle BDC(2), (2) Prop.54 therefore BF is also double of the triangle BAC. Schol. Hence it is evident, that a parallelogram is double of a triangle, if they have equal bases, and be between the same parallels.

Fig. 61.

Cor. If a triangle BAC and a parallelogram DF be between the same parallels, and the base BC of the triangle be double that of the parallelogram DC, the triangle is equal to the parallelogram.

Draw AD. Because the triangles BAD and DAC (1)Prop.38 are equal(1), BAC is double of the triangle DAC, but (2) Prop.41 the parallelogram DF is double of the same DAC (2). (3) Ax. 6. therefore the triangle BAC and the parallelogram DF are equal (3).

Fig. 61.

PROP. XLII. PROB.

To construct a parallelogram equal to a given triangle (BAC), and having an angle equal to a given one (0).

Through the point A draw the right line AF parallel

to BC, bisect BC the base of the triangle in D, and at the point D and with the right line CD make the angle CDE equal to the given one O, through C draw ČF parallel to DE, until it meet the line AF in F; DF is the required parallelogram.

Because EF is parallel to DC(1), and CF parallel to (1) Constr. DE(1), DEFC is a parallelogram (2), and has the angle (2)Def. 50: CDE equal to the given one O (1); and it is equal to

the triangle BAC, because it is between the same (3) Cor. parallels and on half the base of the triangle (3).

PROP. XLIII. THEOR.

Prop. 41.

In a parallelogram (AC) the complements (AK and Fig. 62. KC) of the parallelograms about the diagonal (FH and GE) are equal.

Draw the diagonal DB and through any point in it K, draw the right lines FE and GH parallel to AB and BC, then FH and GE are the parallelograms about the diagonal, AK and KC their complements.

Because the triangles BAD and BCD are equal (1), (1) Prop.54 and the triangles BGK, KFD are equal to BEK,KÈD (1), take away the equals BGK and KFD, BEK and KHD from the equals BAD and BCD, and the remainders, namely, the complements AK and KC are equal.

Cor. The parallelograms OF and EK about the Fig. 65. diagonal of a square AD are squares.

Because the triangle BAC is isosceles, and the angle

at A right, ABC is half a right angle (1), since, then, (1) Cor. 3. in the triangle OBG, the angle at O is right for it is Prop.32. external to a right angle (2), and the angle ŎBG a half (2) Prop.29 right angle, the angle OGB must also be half a right angle, and therefore the sides OG and OB are equal

(3); it is evident therefore, that OF is a square. In (3) Prop.6. the same manner it can be demonstrated that EK is a square.