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Moreover, since it was shown that the two parts BAD and CAD of the < BAC are equal respectively to the two parts, A D C and B D A, of the < BDC, if the one pair of equal <s be added to the other the sums will be equal; that is to say, the < BAC is equal to the < BDC.
It has thus been shown that in the parallelogram ABDC the opposite sides are equal—namely, A B to CD, and AC to BD; and that the opposite s are equal, namely, the Z A B D to the < ACD, and the <BAC to the < BDC; and that the diagonal AD divides the parallelogram into two As which are equal in area.
PROPOSITION XXXV. Parallelograms upon the same base and between the same parallels are equal to one another.
For the proof of this proposition we require to
know,1. That if two straight lines are parallel and are
intersected by a third straight line, the exterior < 80 formed is equal to the interior and opposite < on the same side of the intersecting line.
(Prop. XXIX.) 2. That if two As have two Zs of the one equal
to two Zs of the other, each to each, and likewise a side of the first equal to a side of the second, similarly placed with respect to the equal <s, those s will also be equal in every
other respect. (Prop. XXVI.) 3. That if equal magnitudes be taken from the
same magnitude, the magnitude remaining in the one case will be equal to the magnitudo remaining in the other. (Ax. III.)
(The following demonstration is applicable to any one of the three figures given, which indicate the different positions which the parallelograms may occupy with relation to each other, with this exception, that in applying the demonstration to the first figure, wherever the letter E occurs, the letter D must be substituted for it. The second and third figures had better be employed first in going through the proposition. The peculiarity of the first figure is that the points D and E coincide.)
Let A BDC, and ABFE be two parallelograms upon the same base, A B, and between the same parallels, A B and CF. It has to be shown that they are equal in area.
To demonstrate this it is shown: 1st. That the As CAE and DBF are equal in area. 2nd. That the remainder left, when the A CAE is taken from the trapezium C AB F, is equal to the remainder left when the ABDF is taken away from the same figure.
Since the straight lines CA and D B are parallel and the straight line C F intersects them, the exterior <FDB is equal to the interior and opposite < on the same side of the intersecting line, namely, ECA. (Prop. XXIX.)
Again, since the straight lines AE and B F are parallel, and the straight line C F intersects them, the exterior < CEA is equal to the interior and opposite < on the same side of the intersecting line-namely, DFB.
And because C ABD is a parallelogram, its opposite sides A C and B D are equal.
Hence it appears that the two As CAE and DBF
have two Zs of the one (namely ACE and A E C) equal respectively to two Zs of the other (namely BDF and BFD), and a side of the one (viz. A C) equal to a side of the other similarly placed with respect to the equal 28 (namely BD).
These As are therefore equal in every other respect; and, among these respects, in area.
Now if the ACA E be taken away from the figure A B.FC, we have left the parallelogram A BFE; and if the ADBF be taken away from the same figure ABFC, we have left the parallelogram A BD C.
But since the As are equal in area, the remainder in the first case must be equal to the remainder in the second. That is to say, the parallelogram ABF E is equal in area to the parallelogram ABDC.
Let this proposition next be gone through with the figures drawn in an inverted position, and then with the use of different letters.
The altitude of a parallelogram is the length of a perpendicular drawn from the base on which it stands to the
opposite side, or the opposite side produced.
It follows from the 28th proposition that perpendiculars to the same right line are parallel—the two interior angles being together equal to two right Zs.
It follows accordingly from the 33rd proposition that if two perpendiculars to the same right line are of equal length, the line joining their extremities is parallel to the line on which they stand.
It also follows from the 34th proposition that all perpendiculars drawn between two parallel straight lines are of the same length.
Hence it appears that if two parallelograms are between the same parallels, they may also be described as having the same altitude. And vice verså, that if two parallelograms have their bases in the same straight line, and have the same altitude, they may also be described as being between the same parallels.
Parallelograms upon equal bases and between the same parallels (or having the same altitude) are equal to one another.
For the construction in this proposition we must
be able to join two given points by a straight
line. For the demonstration of the proposition we must
know,1. That the opposite sides of parallelograms are
equal. (Prop. XXXIV.) 2. That the lines joining the corresponding ex
tremities of equal and parallel straight lines are themselves equal and parallel. (Prop. XXXIII.) 3. That parallelograms upon the same base and
between the same parallels are equal. (Prop.
XXXV.) 4. That things that are equal to the same are
equal to each other. (Ax. I.) Let ABCD and EF GH be two parallelograms
upon equal bases (A B and G
EF), and between the same parallels. It has to be shown that they are equal to each other
Join the points A and
H by the straight line A H, and the points B and G by the straight line B G.
We thus get a quadrilateral figure A BGH, which it is easy to show is a parallelogram. For, since the figure EFGH is a parallelogram, the opposite sides EF and G H are equal,
Now E F is by hypothesis equal to A B.
Moreover, the lines D G and AF are supposed to be parallel to start with.
Consequently, the lines A B and H G are both equal and parallel
It follows, therefore (according to the 33rd proposition), that the lines A H and BG, which join their corresponding extremities, are also parallel; so that the figure A BGH is a parallelogram.
Now the parallelograms A B C D and A BGH are upon the same base, A B, and between the same parallels, A B and DG. They are therefore equal in area. (Prop. XXXV.)
Again, the parallelograms HGBA, and HGFE are upon the same base, HG, and between the same parallels, HG and A F. They are therefore equal in area. (Prop. XXXV.)
But things that are equal to the same are equal to each other. (Ax. I.)
Hence the parallelograms ABCD and EFGH, being each equal to the parallelogram A BGH, are equal to each other,
PROPOSITION XXXVII. Triangles upon the same base and between the same parallels are equal to each other.
For the construction in this proposition we must
be ableto draw a straight line from a given point, so as to be parallel to a given straight
line. (Prop. XXXI.) To prove the proposition by the aid of the con
struction we must know, 1. That a parallelogram is divided into two equal
parts by either of its diagonals. (Prop. XXXIV.)