or, which is equal [3. 2], to the rectangle CDB, with the square of DB; therefore the rectangle ADB is equal to double the rectangle CDB, with the square of DB; adding to each the square of CD, the rectangle under AD, DB with the square of CD is equal to double the rectangle CDB with the squares of CD, DB, or, which is equal [4. 2], to the square of CB.

Cor. 1.-Hence, the rectangle under the parts of any right line divided into two parts, is greatest, when the right line is bisected, as in C; and, the nearer the point of division is to the middle of the line, the greater the rectangle; the difference between it and the square of the half line, being [by this prop.] the square of the intermediate part CD.

Cor. 2.-If from the extremes [A, B], of any right line [AB], equal parts [AE, DB] be taken, and any point [F] be taken between them; the rectangle [AFB], under the segments [AF, FB] of the whole line [AB] between its extremes and the last assumed point, is equal to a rectangle [ADB], under the segments [AD, DB] of the same whole line between its extremes and one of the first assumed points, with the rectangle [EFD] under the segments between the last assumed point [F], and those [E, D], which are equidistant from the extremes of the whole.

A E

C

F D B

Bisect AB in C.

The rectangle AFB with the square of CF, is equal to the square of CB [5. 2]; or, which is equal [by the same], to the rectangle ADB with the square of CD; or, the square of CD being equal to the rectangle EFD with the square of CF [5. 2], to the rectangles ADB, EFD and the square of CF; taking from each the common square of CF, there remains the rectan gle AFB, equal to the rectangles ADB, EFD.

PROP. VI. THEOR.

If a right line (AB) be bisected, and another right line (BD) be added in continuation; the rectangle (ADB) under the whole compound (AD) and the part added (BD), with the square of the half line (CB), is equal to the square of the compound of the half line and part added (CD),

On CD describe the square CDFE [46. 1], draw DE, through B, draw BG parallel to DF, meeting ED in H; and through H, draw MK parallel to AD, meeting a right line, drawn through A parallel to CE, in K.

Because AC, CB are equal, the rectangles AL, CH are equal [36. 1], but CH, HF. being complements, are equal [43. 1], therefore the rectangles AL, HF are equal [Ax. 1. 1]; adding to each the rectangle CM, the rectangle AM is equal to the gnomon CMG, and adding to each of these LG, the rectangles AM, LG are equal to the square CF of CD.

But AM is the rectangle under AD and BD, for DM is equal to DB [Cor. 2. 46. 1. and Def. 36. 1]; and LG is the square of LH [Cor. 2. 46. 1], and therefore of CB, which is equal to LH [34. 1].

Otherwise.

On the given right line DA produced beyond A, take AZ equal to BD [3. 1], adding to each AB, ZB is equal to AD; and ZD is divided equally in C, and unequally in B, therefore the rectangle ZBD with the square of CB, is equal to the square of CD [5. 2], and therefore, the rectangle ZBD_being_equal to the rectangle ADB, because ZA is equal to BD and ZB to AD, the rectangle ADB with the square of CB, is equal to the square of CD.

Scholium.-A like observation, as is made in the scholium to prop. 3 of this book, is applicable to this proposition and the preceding, which may be virtually comprised in one proposition, namely, The difference of the squares of two right lines, is equal to the rectangle under their sum and difference;" as is manifest, by supposing, in both propositions, the right lines to be AC, CD.

Theorem.-If the excess of the first [AB] of four magnitudes [AB, CD, EF, GH above the second [CD], be equal to the excess of the third [EF] above the fourth [GH], the first [AB] being greater than the third [EF]; the excess of the first [AB], above the third [EF], is equal to the excess of the second [CD] above the fourth (GH).

A

K

-B

D

L

E

-F

-H

G

Let AK be a part taken on AB equal to CCD, and EL a part on EF equal to GH; and, KB, LF are the excesses of AB above CD, and of EF above GH, and are equal (Hyp.); whence the excess of AB above EF, being equal to the excess of AB above EL and LF together, or above their equals GH and KB together; taking from each KB, the excess of AB above EF, is equal to the excess of AK, or its equal CD, above GH.

Corollaries to the two preceding propositions.

Cor. 1.-If a perpendicular (CD) be let fall on the base (AB) of a triangle, from the opposite angle (ACB); the rectangle under the sum and difference of the sides [AC, CB), is equal to the rectangle under the sum and difference of the segments (AD, DB) of the base, intercepted between its extremes (A, B), and the perpendicular (CD).

Let AC, see both figures to this cor., be that side, which is not the less of the two; and the difference of the squares of AC and AD, and of BC and BD are equal, being each equal to the square of CD (47. 1); therefore by (Theor. at this prop.) the difference of the squares of AC and CB, or, which is equal (Schol, to this prop.), the rectangle under the sum and difference

of AC and CB, is equal to the difference of the squares of AD and BD, or, which is equal (Schol. to this prop.), the rectangle under the sum and difference of AD and DB,

Cor. 2.--To divide a given right line (AB) so into two parts, that the rectangle under the parts, may be equal to the

A

D

СЕ B

square of a given right line (D), not greater than the half (CB) of the first mentioned right line.

On CB, take CE, whose square is equal to the excess of the square of CB above that of D (Cor. 2. 47. 1); and, since the square of CP is equal to the two squares of CE and D (Constr.), taking from each the square of CE, there remains the square of D equal to the excess of the square of CB above that of CE, or, which is equal [5. 2], the rectangle AEB.

Cor. 3.-To add to a given right line (AB) such a part, that the rectangle under the whole and part added, may be equal to the square of a given right line (D).

Bisect AB in C, and on CB produced take CE, whose square is equal to the squares of CB and D together (Cor. 1. 47. 1); and, since the square of CE is equal to the two squares of CB and D together (Constr.), taking from each the square of CB, there remains the square of D equal to the excess of the square of CE above that of CB, or which is equal (6. 2), the rectangle AEB.

Cor. 4.-In equiangular triangles (ABC, DEF, having the angles at A and D equal, as also those at B and E, and at C and F), the rectangles under the sides about any of the equal angles (as ABC and E) taken alternately, are equal, (namely, the rectangles, formed of sides, taken from different triangles and opposed to contrary angles, as under AB, EF, and under BC, DE).

In AB produced, take BG equal to EF, and in CB produced, BH equal to DE, and join GH; and, because the angle HBG is equal to ABC (15. 1), or its equal (Hyp.) E, and the sides BG, BH severally equal to EF, ED (Constr.), the triangle HBG is equiangular to the triangle DEF (4. 1), and therefore to the triangle ABC (Hyp. and Ax. 1. 1), having the angles at A and H equal, and at C and G; join CG, bisect AC, CG in K and L, by the perpendiculars KM, LM (10. and 11. 1), meeting each

other in M.

Join MA, MG, and draw MN perpendicular to AG (12. 1), and, because ML, LG and the angle MLG, are severally equal to ML, LC and the angle MLC, the right lines MG, MC are equal (4. 1); in like manner MA, MC may be proved equal; and, because the triangle MCG is isosceles, and MA equal to MC, the angle CAG is half of the angle CMG (Cor. 3. 32. 1); and, since CHG has been proved equal to CAG, the angle CHG is also half of CMG, therefore MH is equal to MC (Cor. 4. 32. 1); and, in the isosceles triangle AMG, the perpendicular MN bisects the base AG in N (Cor. 26. 1); and, in the triangle BMG, the rectangle under the sum and difference of MG and MB, or, BH being equal to the sum, and CB to the difference, of these right lines, the rectangle CBH, is equal to the rectangle under the sum and difference of GN, BN, or the rectangle ABG (Cor. 1. 5 and 6. 2); and BH is equal to DE, and BG to to EF, therefore the rectangle under BC and DE, is equal to the rectangle under AB and EF.