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On A B describe the ABD E.
or BD, intersecting A D in G and
Through G draw KGH parallel to A B or ED.
It is self-evident that the whole O AD is equal to the sum of the four rectangles KC, GB, E G and FH, for they are parts into which it is divided. But KC and FH are as (Prop. A.), and KC is the
on A C, and FH is the on CB, for it is the on GH which is equal to C B. (I. 34.)
E G is the rectangle contained by A C and C B, for it is contained by K G, which is equal to A C (I. 34) and FG, which is equal to FD, which is equal to CB. (I. 31.)
GB is also the rectangle contained by A C and CB, for it is contained by GC and C B, and GO and AO are equal because they are sides of the same square.
Consequently EG and G B are together equal to twice the rectangle* contained by A C and C B.
We have thus shown that the on the whole line (A B) is equal to the sum of the Os on the two parts (A C and C B) together with twice the rectangle under the two parts.
The beginner will observe that the stress of the proof does not consist in showing that the whole figure AD is
* It would do equally well to point out that the rectangles E G and G B are equal, being complements of the parallelograms K C and F H about the diagonal, and that G B is the rectangle contained by AC and CB; so that EG and G B together are equal to twice the rectangle contained by A C and CB.
equal to the sum of the parts into which it is divided, for that is self-evident, but in showing that the whole figure and its parts answer to the description given of them in the enunciation.
If a right line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
For the construction employed we must be able 1. On a given straight line to describe a square.
(I. 46.) 2. Through a given point to draw a line parallel
to a given right line. (I. 31.) We must also know,1. That if equals are added to equals, the sums
are equal. (Ax. II.) 2. That the opposite sides of a parallelogram are
equal. (I. 34.) 3. That the parallelograms about the diagonal of
a square are themselves squares. (Prop. A.) 4. That the complements of the parallelograms
about the diagonal of a parallelogram are equal.
(I. 43.) 5. That parallelograms upon equal bases and between the same parallels are equal. (I. 36.)
Let the line A B be divided into two equal parts in C, and into two L unequal parts in D.
We have to prove that rect. AD,
D B, together with the oon CD, is equal to the on C B.
On C B describe the DCBEF. Draw the diagonal FB.*
Through D draw the right line D G parallel to CF or B E, intersecting F B in H, and FE in G.
Through H draw L K parallel to B A, and through A draw A K parallel to CF or BE, and meeting KL in the point K.
1. Because AC and C B are equal, A O and C L are parallelograms standing on equal bases and between the same parallels; therefore the rectangle A O is equal to the rectangle C L.
2. The rectangles CH and HE are equal, because they are the complements of the parallelograms OG and D L, which are about the diagonal F B.,
3. If the equal rectangles CH and H E be added respectively to the equal rectangles A O and CL, the sums will be equal; that is to say, the rectangle AH. (which is the sum of A 0 and C H) will be equal to the gnomon CLG (which is the sum of C L and G L).
4. If to each of these equals the OOG be added, the sums will be equal; that is to say, the sum of the rect. A H and the O O G will be equal to the sum of the gnomon CLG and the OOG.
5. But A H is the rectangle contained by A D and DB, for it is contained by AD and DH, and DH and DB are equal, being sides of the same 0; and OG is the on CD, for it is the on OH which is equal to C D. (I. 34.) 6. Therefore the rect. A D, D B together with the
on CD is equal to the gnomon CLG together with the O OG.
* Let the beginner take particular notice that the diagonal needed is the one which is drawn to the extremity of the line A B. The diagonal CE would not do.
7. But the gnomon CLG together with the OUG is equal to the CE, which is the on CB.
Therefore the rectangle under the unequal parts, A D and D B, together with the oon ( D, the part between the points of section, is equal to the on CB, which is half the line A B.
The steps of this proof may be easily indicated by the use of the following equations, which express what is stated in the preceding paragraphs which bear the same numbers :
1. Rect. AO = rect. OL. 2. Rect. UH= rect. H E.
3. Therefore rect. AO + rect. CH = rect. CL + rect. HE, or rect. AH = gnomon CLG.
4. Add to both the OOG, then rect. AH +0 OG = = gnomon CLG +0 0 G.
5. But A H = rect. A D, D B, and O G = CD.
6. Therefore rect. AD, DB+ on CD= gnomon CLGE OOG.
7. But gnomon CLG +OOG = on C B.
8. Therefore rect. AD, DB + on CD = = on СВ.
If A C and CD be regarded as two separate lines, it is clear that A D is the sum of the two, and D B the difference between them, for D B is what is left when C D is taken from C B, and C B is equal to A C. Therefore A H is the rectangle under the sum and the difference of the two lines A C and CD. OG is the on the smaller of the two lines; and C E is the on the larger, for it is the O on CB, which is equal to A C.
Consequently what is proved in this proposition may be thus stated: “The rectangle contained by the sum and the difference of two unequal lines, together with the o on the smaller, is equal to the on the larger;” or thus : " The difference between the Os on two unequal lines (as AC and CD) is equal to the rectangle contained by their sum and their difference. This is incomparably the most useful form in which the proposition can be stated.
If a right line be bisected and produced to any length, the rectangle under the whole line thus produced and the produced part, together with the square on half the line, is equal to the square on the line made up of the half line and the produced part.
The propositions required for the construction and proof in this proposition are the same as in the last.
Let A B be a right line bisected in C, and produced to D. We
have to prove that the rect. A D, R DB, together with the on CB, is equal to the on CD.
On CD describe the OCDEF. Draw the diagonal F D (see the note on the last proposition.)
Through B draw BG parallel to CF or D Е, cutting FD in H and F E in G.
Through H draw KHM parallel to A D, and through A draw AM parallel to D Е, and meeting KM in the point M.
1. The rectangles A L and C H are on equal bases
* The beginner will remember that the difference between two magnitudes is what is added to the smaller to make a quantity equal to the larger; and the rectangle A H is added to the DOG to make a quantity equal to the o CE.