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Base PT, and so is MC to CT. , Therefore as the
Base EH is to the Base NP, fo is MC to CT; but
CT is equal to AG. Wherefore as the Base E H is
to the Base NP, so is MC to AG. Therefore the i
Bases and Altitudes of the equal solid Parallelepipedons
AB, CD, are reciprocally proportional.

Now, let the Bases and Altitudes of the folid Pa-
rallelepipedons AB, CD, be reciprocally proportion
al; that is, let the Base EH be to the Base N P;.
as the Altitude of the Solid CD is to the Altitude of
the Solid AB. I say; the Solid AB is equal to the
Solid CD.

For let again the insistent Lines be at Right Angles
to the Bases; then if the Base EH be equal to the
Base NP, and E H is to NP, as the Altitude of the
Solid CD is to the Altitude of the Solid AB; the
Altitude of the Solid .C D shall be equal to the Alti-
tude of the Solid AB. But solid Parallelepipedons
that stand upon equal Bases, and have the fame Alti-
tude, are * equal to each other. Therefore the Solid * 31 of thise
AB is equal to the Solid CD.

But now let the Base E H not be equal to the Base NP; and let E H be the greater ; then the Altitude of the Solid CD is greater than the Altitude of the Solid AB; that is CM is greater than AG. Again, put CT equal to AG, and complete the Solid CV as before: And then because the Base E H is to the Base NP, as MC is to AG, and AG is equal to CT; it shall be as the Base EH is to the Base NP, fo is MC to CT; but as the Base EH is to the Base NP, so is the Solid AB to the Solid VC; for the Solids AB, CV, have equal Altitudes : And as MC is to CT, so is the Base MP to the Base PT, and so the Solid CD to the Solid CV. Therefore as the Solid AB is to the Solid CV, so is the Solid CD to the Solid CV: But since each of the Solids A B, CD, has the same Proportion to CV, the Solid AB shall be equal to the Solid CD; which was to be demonstrated.

Now let the infiftent Lines FE; BL, GA, KH, XN, DO, MC, RP, not be at Right Angles to the Bases; and from the Points F, G, B, K, X, M, D, Ř, let there be drawn Perpendiculars to the Planes of the Bases EH, NP, meeting the same in the

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Points S, T, Y, V, Q, Z, , , and complete the Solids F V, X2, Then, I say, if the Solids AB, CD, be equal, their Bases and Altitudes are reciprocally proportional, viz. as the Base E H is to the Base NP, so is the Altitude of the Solid CD to the Altitude of the Solid AB.

For because the Solid AB is equal to the Solid * 30 of this. CD, and the Solid AB is * equal to the Solid BT;

for they stand upon the fame Base, have the fame Altitude, and their infiftent Lines are not in the same Right Lines, and the Solid D.C is * equal to the Solid DZ, since they stand upon the fame Base, XR have the same Altitude, and their infiftent Lines are not in the same Right Lines; the Solid BT shall be equal to the Solid DZ; but the Bases and Altitudes

of those equal Solids, whose Altitudes are at Right + From Angles to their Bases, are † reciprocally proportional. wbat bas

Therefore as the Base FK is to the Bale XR, so is proved.

the Altitude of the Solid DZ, to the Altitude of the Solid BT; but the Base F K is equal to the Base EH, and the Base XR to the Base NP. Wherefore as the Base EH is to the Base NP, fo is the Altitude of the Solid DZ to the Altitude of the Solid BT, but the Solids DZ, DC, have the same Altitude, and so have the Solids B T, BA. Therefore the Base EH is to the Bafe NP, as the Altitude of the Solid DC is to the Altitude of the Solid AB; and so the Bases and Altitudes of equal Solids are reciprocally proportional.

Again, let the Bases and Altitudes of the folid Parallelepipedons AB, CD, be reciprocally proportional, viz. as the Base E H is to the Bafe NP, fo let the Altitude of the Solid CD be to the Altitude of the Solid A B. I say, the Solid AB is equal to the Solid CD.

For the fame Construction remaining, because the Bale E H is to the Base NP, as the Altitude of the Solid CD is to the Altitude of the Solid AB; and since the Base E H is equal to the Base FK, and NP to XR, it shall be as the Bafe FK is to the Base XR, so is the Altitude of the Solid CD to the Altitude of the solid AB; but the Altitudes of the Solids A B, BT, are the fame; as also of the Solids CD, DZ. Therefore the Base FK is to the Base XR, as the

Altitude

Altitude of the Solid DZ is to the Altitude of the Solid BT; wherefore the Bases and Altitudes of the folid Parallelepipedons BT, DZ, are reciprocally proportional; but those solid Parallelepipedons, whose Altitudes are at Right Angles to their Bases, and the Bases and Altitudes are reciprocally proportional, are equal to each other. But the Solid B T is equal to the Solid B A; for they stand upon the fame Base FK, have the same Altitude, and their insistent Lines are not in the fame Right Lines, and the Solid DZ is also equal to the Solid DC, since they stand upon the fame Base XR, have the same Altitude, and their insistent Lines are not in the fame Right Lines. Therefore the Solid A B is equal to the Solid CD; which was to be demonstrated.

PROPOSITION XXXV.

THEOREM.

If there be two plane
Angles equal, and

from the Vertices of those Angles two Right Lines be elevated above the Planes, in which the Angles are, containing equal Angles with the Lines first given, each to its correspondent one ; and if in those elevated Lines any Points be taken, from which Lines be draton perpendicular to the Planes in which the Angles first given are, and Right Lines be drawn to the Angles first given from the Points made by the Perpendiculars in the Planes, those Right Lines will contain equal Angles with the elevated Lines.

>Angles ; ånd from A, D, the Vertices of those Angles, let two Right Lines AG, DM, be elevated above the Planes of the faid Angles, making equal Angles with the Lines first given, each to its correspondent one, viz. the Angle MDE equal to the Angle GAB, and the Angle MDF to the Angle GAC; and take any Points G and M in the Right Lines AG, DM, from which let GL, MN, be drawn perpendicular to the Planes passing thro' BAC, EDF, meeting the same in the Points L, N, and

Q2

join

join LA, ND. I say the Angle GAL is equal to the Angle MDN.

Make A H equal to DM, and thro' H let HK be drawn parallel to GL; but GL is perpendicular to

the Plane passing thro' BAC. Therefore HK shall * 8 of this. be * also perpendicular to the Plane passing thro' BAC.

Draw from the Points K, N, to the Right Lines AB,
AC, DE, DF, the Perpendiculars KB, KC, NE,

NF, and join HC, CB, MF, FE. Then because + 47. I. the Square of HA is equal to the Squares of HK,

KA, and the Squares of KC, CA, are + equal to the Square of KĀ; the Square of HA shall be equal to the Squares of HK, KC, ÇA; but the Square of HC is equal to the Squares of HK, KC. Therefore

the Square of HA will be equal to the Squares of 45. 1. HC and CA; and so the Angle HCA is I a Right

Angle. For the same Reason, the Angle D F M is
also a Right Angle. Therefore the Angle ACH is
equal to DFM; but the Angle HAC is also equal
to the Angle MDF. Therefore the two Triangles
MDF, HAC, have two Angles of the one equal to
two Angles of the other, each to each, and one side
of the one equal to one Side of the other, viz. that
which is subtended by one of the equal Angles; that
is, the Side HA equal to DM; and so the other
Sides of the one, shall be * equal to the other Sides
of the other, each to each. Wherefore AC is equal
to DF. In like manner we demonstrate, that AB
is equal to DE; for let HB, ME, be joined. Then
because the Square of AH is equal to the Squares of
AK and KH; and the Squares of AB, BK, are
equal to the Square of AK; the Squares of AB, BK, 11
KH, will be equal to the Square of AH; but the
Square of BH is equal to the Squares of BK, KH;
for the Angle HKB is a Right Angle, because HK
is perpendicular to the Plane passing through B A C.

Therefore the Square of AH is equal to the Squares 0 $43. 1, of AB, BH. Wherefore the Angle AB H is † a Right

Angle. For the fame Reason, the Angle DEM is also a Right Angle. And the Angle BAH is equal to the Angle EDM, for so it is put; and AH is equal to D M. Therefore A B is † also equal to DE. And so fince AC is equal to DF, and AB to DE, the two Sides CA, A B, shall be equal to

the

* 26. 1,

2

the two Sides FD, DE; but the Angle BAC is equal to the Angle FDE. Therefore the Bale BC is * equal to the Base E F, the Triangle to the Tri- * 4. I. angle, and the other Angles to the other Angles. Wherefore the Angle ACB is equal to the Angle DFE; but the Right Angle A CK is equal to the Right Angle DFN; and therefore the remaining Angle B CK is equal to the remaining Angle EFN. For the same Reason, the Angle CBK is equal to the Angle FEN; and so because BCK, EFN, are two Triangles, having two Angles equal to two Angles, each to each, and one Side equal to one side, which is at the equal Angles, viz, BC equal to EF; therefore they shall have the other Sides equal to the other Sides. Therefore CK is equal to FN, but AC is equal to DF. Therefore the two Sides AC, CK, are equal to the two Sides DF, FN, and they contain Right Angles; consequently the Base AK is equal to the Bafe DN. And since AH is equal to DM, the Square of A H shall be equal to the Square of DM; but the Squares of A K, KH, are equal to the Square of AH; for the Angle A KH is a Right Angle, and the Squares DN, NM, are equal to the Square of DM, since the Angle DNM is a Right Angle. Therefore the Squares of AK, KH, are equal to the Squares of DN, NM; of which the Square of A K is equal to the Square of DN. Wherefore the Square of KH remaining, is equal to the remaining Square of NM; and so the Right Line HK is equal to MN. And since the two Sides H A, AK, are equal to the two Sides MD, DN, each to each, and the Base HK has been proved equal to the Base NM, the Angle HA K fall be f equal to the Angle † 8. le MDN; which was to be demonstrated.

Coroll. From hence it is manifeft, that if there be two

Right-lined plane Angles equal, from whose Points
equal Right Lines be elevated on the Planes of the
Angles, containing equal Angles with the Lines first
given, each to each ; Perpendiculars drawn from the
extreme Points of those elevated Lines to the Planes
of the Angles first given, are equal to one another.

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