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*0.5.

of the solids EM, GN. And because AB is to CD, as EF to ST, and that from AB, CD, the solid parallelopipeds AK, CL are similarly described; and in like manner the solids EM, SV, from the straight lines EF, ST; therefore AK is to CL, as EM to SV; but, by the hypothesis, AK is to CL, as EM to GN; therefore GN is equal* to SV: but it is likewise similar and similarly situated to SV; therefore the planes which contain the solids GN, SV are similar and equal, and their homologous sides GH, ST equal to one another: and because as AB to CD, so EF to ST, and that ST is equal to GH, therefore AB is to CD, as EF to GH. Therefore, if four straight lines, &c. Q. E. D.

See N.

PROPOSITION XXXVIII.

THEOR." If a plane be perpendicular to another plane, and " a straight line be drawn from a point in one of the planes " perpendicular to the other plane, this straight line shall "fall on the common section of the planes."

"Let the plane CD be perpendicular to the plane AB, and "let AD be the common section: if any point E be taken in "the plane CD, the perpendicular drawn from E to the plane "AB, shall fall on AD.

"For if it does not, let it, if possible, fall elsewhere, as "EF; and let it meet the plane AB in the point F; and " from F, draw *, in the plane AB, a per

12. 1.

• 3 Def. 11.

† 17.1.

"pendicular FG to DA, which is also per* 4 Def. 11. "pendicular * to the plane CD; and join "EG. Then, because FG is perpendicular "to the plane CD, and the straight line "EG which is in that plane, meets it, "therefore FGE is a right angle*: but EF is also at right "angles to the plane AB, and therefore EFG is a right "angle: wherefore two of the angles of the triangle EFG " are equal together to two right angles; which is † absurd; "therefore the perpendicular from the point E to the plane "AB, does not fall elsewhere than upon the straight line AD; "it therefore falls upon it. If, therefore, a plane, &c. Q. E. D."

C

E

G

A

D

F

B

PROPOSITION XXXIX.

THEOR.-In a solid parallelopiped, if the sides of two of the See N. opposite planes be divided, each into two equal parts, the common section of the planes passing through the points of division, and the diameter of the solid parallelopiped, cut each other into two equal parts.

Let the sides of the opposite planes CF, AH of the solid parallelopiped AF, be divided each into two equal parts in the points K, L, M, N; X, O, P, R; and join KL, MN, XO, PR: and because DK, CL are equal and parallel, KL is parallel * to DC: for the same reason, MN is parallel to BA: and BA is parallel to DC; therefore,

33. 1.

[blocks in formation]

plane. Let YS be the common section of the planes KN, XR; and DG the diameter of the solid parallelopiped AF: YS and DG shall meet, and cut one another into two equal parts. Join DY, YE, BS, SG. Because DX is parallel to OE, the alternate angles DXY, YOE are equal * to one another: and 29.1. because DX is equal to OE, and XY to YO, and that they contain equal angles, the base DY is equal to the base YE, 4.1. and the other angles are equal; therefore the angle XYD is equal to the angle OYE, and DYE is a straight line: for the 14.1. same reason, BSG is a straight line, and BS equal to SG. And because CA is equal and parallel to DB, and also equal and parallel to EG, therefore DB is equal and parallel to 9.11.

EG: and DE, BG join their extremities; therefore DE is equal and parallel * to BG: and DG, YS are drawn from points in the one, to points in the other, and are therefore

33. 1.

in one plane: whence it is manifest, that DG, YS must meet

one another: let them meet in T. And because DE is parallel to BG, the alternate angles EDT, BGT are* equal: and the angle DTY is equal * to the angle GTS: therefore in the

29.1. 15. 1. & 22. 5.

* 21.3.

31.3.

* 4.6.

DE is not greater than its half, but BH taken from AB is greater than its half, therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA, therefore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C.

Q. E. D.

And if only the halves be taken away, the same thing may, in the same way, be demonstrated.

PROPOSITION I.

THEOREM.-Similar polygons inscribed in circles, are to one another as the squares of their diameters.

Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHKL: and let BM, GN be the diameters of the circles: as the square of BM is to the square of GN, so shall the polygon ABCDE be to the polygon

FGHKL.

Join BE, AM, GL, FN: and because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and as BA to AE, so is GF to FL: therefore

the two triangles BAE, GFL,

B

F

EG

L

M

N

C

H

K

having one angle in one, equal
to one angle in the other, and the
sides about the equal angles pro-
portionals, are equiangular; and
therefore the angle AEB is equal
to the angle FLG: but AEB is
equal* to AMB, because they stand upon the same circum-
ference: and the angle FLG is, for the same reason, equal to
the angle FNG; therefore also the angle AMB is equal to
FNG: and the right angle BAM is equal to the right * angle
GFN; wherefore the remaining angles in the triangles ABм,
FGN are equal, and they are equiangular to one another:
therefore as BM to GN, so* is BA to GF; and therefore the

*10 Def. 5. duplicate ratio of BM to GN, is the same * with the duplicate ratio of BA to GF: but the ratio of the square of BM to the square of GN, is the duplicate * ratio of that which BM has to GN; and the ratio of the polygon ABCDE to the polygon

* 20.6.

FGHKL is the duplicate * of that which BA has to GF: therefore as the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore, similar polygons, &c. Q. E. D.

PROPOSITION II.

* 20.6.

THEOR.-Circles are to one another as the squares of their See N. diameters.

Let ABCD, EFGH be two circles, and BD, FH their diameters: as the square of BD to the square of FH, so shall the circle ABCD be to the circle EFGH.

For if it be not so, the square of BD n.ust be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it . First let it be to a space s less than the circle EFGH; and in the circle EFGH† † 6.4. describe the square EFGH. This square is greater than half of the circle EFGH; because, if, through the points E, F, G, H, there be drawn tangents to the circle, the square EFGH is half of the square described about the circle: is less than the square described about it

and the circle * 41. 1. therefore the

square EFGH is greater than half of the circle Divide the

[blocks in formation]

FLG, GMH, HNE, is greater than half of the segment of the circle in which it stands; because, if straight lines touching the circle be drawn through the points K, L, M, N, and the parallelograms upon the straight lines EF, FG, GH, HE be completed, each of the triangles EKF, FLG, GMH, HNE, is the half of the parallelogram in which it is: but every segment is less than the parallelogram in which it is: wherefore

41. 1.

† For there is some square equal to the circle ABCD; let P be the side of it, and to three straight lines, BD, FH, and P, there can be a fourth proportional; let this be Q: therefore the squares of these four straight lines are + 22.6. proportionals; that is, to the squares of BD, FH, and the circle ABCD, it is possible there may be a fourth proportional. Let this be S. And in like manner are to be understood some things in some of the following propositions.

* 21.3.

*31.3.

4.6.

DE is not greater than its half, but BH taken from AB is greater than its half, therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA, therefore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C.

Q. E. D.

And if only the halves be taken away, the same thing may, in the same way, be demonstrated.

PROPOSITION I.

THEOREM.-Similar polygons inscribed in circles, are to one

another as the squares of their diameters.

Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHKL: and let BM, GN be the diameters of the circles: as the square of BM is to the square of GN, so shall the polygon ABCDE be to the polygon

FGHKL.

Join BE, AM, GL, FN: and because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and as BA to AE, so is GF to FL: therefore the two triangles BAE, GFL,

having one angle in one, equal
to one angle in the other, and the
sides about the equal angles pro-
portionals, are equiangular; and
therefore the angle AEB is equal
to the angle FLG: but AEB is

B

A

F

EG

L

N

M

T H

K

equal to AMB, because they stand upon the same circumference: and the angle FLG is, for the same reason, equal to the angle FNG; therefore also the angle AMB is equal to FNG: and the right angle BAM is equal to the right angle GFN; wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another: therefore as BM to GN, so is BA to GF; and therefore the

10 Def. 5. duplicate ratio of BM to GN, is the same * with the duplicate ratio of BA to GF: but the ratio of the square of BM to the square of GN, is the duplicate* ratio of that which BM has to GN; and the ratio of the polygon ABCDE to the polygon

& 22. 5.

20.6.

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