Book II. the fquare of EB is equal to the fquares of ED, DB. Therefore the fquares of AB, BC will be equal to twice the squares of A E, EB.

Again [fig. 2.] the fquare of AB [by 13. 2.] exceeds the fquares of AE, E B by twice the rectangle under A E, E D (because the right line A c is bifected, and CD is added to it). And the fquare of B C [by 12. 2.] is less than their fum by twice the rectangle under c E, E D, that is, twice the Jectangle under A E, E D. Therefore if to the fquare of B C be added the excess whereby the square of AB exceeds the fquares of A E, EB; the fquares of A B, B C, will be equal to twice the squares of A E, E B.

Wherefore, &c. Which was to be demonstrated.

B

SCHOLIU M.

This is one of the useful and elegant propofitions of the antients, to be found in Pappus's Mathematical collections, and from it may be easily obtained the following theorem. If the bafe AC of any triangle A B C be divided into any number of equal parts, of which A D, IC are the first and last; and if right lines B D, B 1 be drawn from the angle B oppofite to the bafe joining those two remote points of divifion, the Squares of the fides A B, B C, of the triangle together, are equal to the fquares of the right lines B D, BI, together A Ꭰ with the fquare of one of the parts AD, or IC, of the bafe, as many times taken wanting two, as there are unites in twice the number of parts the bafe is aivided into.

PROP. IV. THEOR.

If the three fides of any triangle be bifected, and right lines be drawn from the oppofite angles to the points of bifection, four times the fum of the Squares of thefe three bifelling right lines will be equal to thrice the fum of the fquares of the fides of the triangle.

Let there be any triangle A B C, whofe three fides are bifected in the points D, E, F, to which are drawn the

right lines A F, B E, C D from the oppofite angles: I fay, four times the fum of the fquares of AF, BE, CD will be equal to thrice the fum of the fquares of the fides A B, A C, CB of the triangle.

For fince [by third of this] twice the fquares of AF, CF are equal to the fquares of the fides AB, AC of the triangle, and twice the fquares of E B A E is

equal to the fquares of the fides

A B, BC of the triangle, and

twice the fquare of c D, A D is

equal to the fquares of the fides

AC, CB of the triangle. There- A

D

E

B

F

C

fore twice the fums of the former fquares, that is, twice the fum of the fquares of A F, CF, E B, A E, CD, AD will be equal to twice the fum of the fquares of A B, C B, A C. And fo the fum of the squares of A F, CF, E B, A E, CD, A D will be equal to the fum of the fquares of A B, C B, A C. And fince [by 4. 2.] the fquare of any line is equal to four times the fquare of one half of it, four times the fum of the fquares of CF, AE, AD is equal to the fum of the fquares of CB, A C, AB; therefore the fum of the fquares of CF, AE, AD will be equal to one fourth part of the fum of the fquares of c B, A C, AB. And fo the fum of the fquares of A F, E B, CD, together with one fourth part of the fum of the fquares of CB, AC, AB, will be equal to the fum of the fquares of C B, A C, A B. And taking away one fourth part of the fum of the fquares of CB, A C, A B from both, the fum of the squares of AF, EB, CD will be equal to three fourth parts of the fum of the fquares of C B, A C, AB. And taking four times each fum, four times the fum of the fquares of A F, E B, CD will be equal to three times the fum of the fquares of c B, A C, A B.

Therefore, &c. Which was to be demonftrated.

PROP. V. THEOR.

In every parallelogram the fquares of the fides taken together, are equal to the fquares of the diagonals.

Let there be any parallelogram A B C D, whose diagonals are AC, DB: I say, the fquares of the fides AB, BC, CD, AD together, will be equal to the two squares of the diagonals

AC, B D.

For let E be the point of interfection of the diagonals. Then because A B, DC are parallel, the alternate angles A BE, EDC, are [by 29. 1.] equal; and because A D, BC are parallel, the alternate angles B A E, ECD are also equal. And fince the fides A B, D C alfo are equal; there are two

A

B

E

D

C

triangles ABE, DEC having two angles of the one refpectively equal to two angles of the other, and the fides between the equal angles alfo equal. Wherefore [by 26. 1.] the other fides will alfo be equal, and the remaining angles. Wherefore B E will be equal to E D, and A E equal to E c. Therefore the diagonals of the parallelogram ABCD do mutually bifect one another. And so because [by 4. of this] the squares of A B, B C are equal to twice the squares of A E, E B, and the fquares of A D, D c equal to twice the squares of A E, ED, therefore the fum of the fquares of A B, B C, A D, DC will be equal to twice the fum of the fquares of AE, BE, AE, ED. That is, equal to four times the fquare of A E, together with twice the fquares of B E, E D. That is [fince BE has been proved to be equal to E D] to four times the square of AE, and four times the fquare of B E. But [by 4. 2.] the fquare of A c is equal to four times the fquare of A E, and the fquare of B D equal to four times the square of B E. Therefore the fquares of the fides AB, BC, CD, AD will be equal to the fquares of the diagonals A C, B D.

Therefore, &c. Which was to be demonftrated.

SCHOLIU M.

The following theorem in a fix-fided parallelogram is likewife true, viz. that the fum of the fquares of the fides is equal to the difference between twice the fum of the fquares of the three diagonals joining the oppofite angles, or interfecting one anoother in one point, and the fum of the fquares of the fix remaining diagonals. There are alfo other fuch like properties of eight-fided parallelograms, &c. but these I leave for the difcovery of others.

PROP.

PROP. VI. THEOR.

In any quadrilateral figure, the fum of the Squares of the four fides will be equal to the fum of the Squares of the diagonals, together with four times the fquare of the right line joining the middle points of the diagonals.

Let there be a quadrilateral figure AFDE, and let its diagonals AD, E F be bifected in the points в, c: I say, the fum of, the fquares of the four fides A F, F D, AE, E D will be equal to the fum of the fquares of the diagonals A D, F E together with four times the fquare of the right line B C joining the points of bifection B, C, of the diagonals.

For join the right lines F B, B E. Then because the base AD of the triangle AFD is bifected in B; the fquares of A F, FD [by 4. of this] are equal to twice the fquares of AB, BF. And likewise the

BE.

fquares of the fides AE, ED of the triangle A E D equal to twice the squares of A B and Therefore the sum of the fquares of AF,FD,AE,ED will be equal to four times the square of AB, together with twice the fquares of FB, BE. Or fince [by 4. 2.] four times the fquare of AB is equal to the fquare of the diagonal AD, the sum of the

F

B

D

C

E

fquares of the four fides of the figure will be equal to the fquare of the diagonal A D, together with twice the squares of F B, BE. But because the fide F E of the triangle F B E is bifected in C, the fquares of F B, BE [by 4. of this] are equal to twice the fquares of Fc, and CB; and fo twice the fquares of FB, BE are equal to four times the square of F c, and four times the fquare of c B. Therefore the fum of the squares of the fides will be equal to the square of the diagonal A D, together with four times the fquare of F C, and four times the fquare of c B; that is, fince the fquare of the diagonal FE is [by 4. 2.] four times the fquare of

H 3

FC,

Fc, the fum of the fquares of the fides AF, FD, AP, ED will be equal to the fum of the fquares of the diagonals A D, E F, together with four times the fquare of the right line B C joining the middle points B, C, of the diagonals. Therefore, &c. Which was to be demonftrated.

LEMM A.

If the two fides of a triangle be bisected, a right line joining the points of bifection will be parallel to the bafe of the triangle, and equal to one balf of it.

Let there be a triangle A B C, whofe fides A B, C B, are bifected in the points D, E: I fay, the right line DE joining the points D, E will be parallel to the base A c of the triangle, and equal to one half of it.

For draw by 31. 1.] BF parallel to A C, and thro' E draw GF parallel to the fide AB meeting the base A c in the point G, and the right line B F in the point F.

B

F

1

Now because [by 15. 1.] the angle B E F is equal to the angle GEC, and the angle EBF [by 29. 1.] is equal to the angle GCE, and the fide BE [by fuppofition] is equal to the fide EC. Therefore [by 26. 1.] the right line BF will be equal to GC. But [by 34. 1.] AG is equal to B F, and A B equal to GF. Wherefore B D will be equal to F E, that is, equal to E G.

D

G

E

C

Again, because B E is equal to E C, and B D equal to E G, and [by 4. 1.] the angle DBE is equal to the angle GEC; therefore [by 4. I.] DE will be equal to GC; that is, equal to BF, or equal to A G, and the angle B ED will be equal to the angle A C B. Therefore [by 28. 1.] DE will be parallel to A C. And it has been proved to be equal to A G or GC. Wherefore DE will also be equal to one half the base A C. Which was to be demonstrated.

PROP.