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.

Secondly, let the point M be without the right line CH, as in

fig. 2. and 3.

Construct as in the former case, join moreover MB, MC, and draw MO perpendicular to CH; and, as before, the triangles ABC, HBG may be proved equiangular, and the right lines MA, MC, MG equal; and, because the triangle CMG is isosceles, and MA equal to MC, the angle CAG is equal to half of CMG (Cor. 3. 32. 1); but CHG is equal to CAG, and therefore also the half of CMG, therefore MH is equal to MC (Cor. 4. 32. 1), and the triangle MCH isosceles, therefore the perpendicular MO bisects the base CH (Cor. 26. 1); and, in the triangle BMH, the rectangle under the sum and difference of BO, OH, or the rectangle CBH, is equal to the rectangle under the sum and difference of BM, MH (Cor. 1. 5 and 6. 2), or of BM, MG, or, which is equal (by the same Cor.), to the rectangle under the sum and difference of BN, NG, or the rectangle ABG; but BH is equal to DE, and BG to EF, therefore the rectangle under BC and DE, is equal to the rectangle under AB and EF.

In like manner it may be proved, that the rectangle under the sides about the equal angles at C and F, or at A and D, of the triangles ABC, DEF, taken alternately, are equal.

PROP. VII. THEOR.

If a right line (AC), be divided into any two parts (AB, BC); the squares of the whole (AC), and one of its parts (BC), are equal to double the rectangle (ACB) under the whole and that part, with the square of the other part (AB).

B

On AC describe the square ACFD [46. 1], draw CD, through B, draw BE parallel to AD, meeting CD in G, and through G, HK parallel to AC.

The square AF is equal to the rectangles AK, KE with the square HE; add to each, the square BK, the squares AF, BK of AC, BC, are equal to the rectangles AK, BF with the square HE.

But the rectangles AK, BF are, each of them, equal to the rectangle ACB, because CK is equal to BC, and CF to AC [Cor. 2. 46. 1. and Def. 36. 1], and HE is the square of HG, or its equal [34. 1] AB; therefore the squares of AC, BC, are equal to double the rectangie ACB with the square of AB.

Otherwise.

The square of AC, is equal to the squares of AB, BC with double the rectangle ABC (4. 2); add to each the square of BC, and the squares of AC, BC, are equal to the square of AB with double the square of BC and double the rectangle ABC; but double the square of BC with double the rectangle ABC, is equal to double the rectangle ACB [3. 2]; therefore the squares

of AC, BC are equal to double the rectangle ACB with the square of AB.

Scholium.-A like observation, as is made in the scholium to the preceding proposition, and 3rd. of this book, is applicable to the 4th proposition and this, that they may be virtually reduced to one, namely; The sum of the squares of two right lines, is less than the square of their sum, and greater than the square of their difference, by double the rectangle under these right lines; as is manifest, by supposing, in both propositions, the right lines to be AC, CB.

Corollaries to the preceding propositions of this book.

H B

Cor. 1.-If a right line (AB) be so A G divided into two points (G, H), that 1the rectangle (AGB) under the distances of one of these points from its extremes, be equal to the rectangle (AHB) under the distances of the other point from the same extremes, the extreme segments (AG, HB) are equal.

For since the rectangles AGB, AHB are equal (Hyp.), taking from each the rectangle under AG and HB, there remain (by 1. of this,) the rectangles AGH, GHB equal to each other; whence, the side GH being common to both rectangles, the other sides AG, HB are equal (40 and 41. 1).

A.

B H

Cor. 2.-If to a right line (AB), G there be added on both extremes, such – parts (AG, BH), that the rectangles (AGB, BHA) under the added parts (AG, BH,) and the compounds (GB, AH) of the same right line and parts added, be equal; the added parts (AG, BH) are equal.

For since the rectangles AGB, AHB are equal (Hyp.), adding to each the rectangle under AG and BH, the totals, which are (by 1. of this,) the rectangles AGH, BHG, are equal; whence, the side GH being common to both rectangles, the other sides AG, BH are equal (40 and 41. 1).

PROP. VIII. THEOR. (See note.)

If a right line (AB), be divided into any two parts (AC, CB); four times the rectangle under the whole (AB), and either part (as CB), with the square of the other part (AC), is equal to the square of the compound of the whole (AB) and part first taken (CB), as of one right line.

On AB produced, take BD equal to CB, and on AD describe the square ADFE [46. 1]; draw DE, through B and C, BL and CH parallel to DF [31. 1], meeting DE in K and P, and through K and P, MN, XO parallel to ÃB.

Because BN, GR are squares of BD, GK (Cor. 2. 46. 1), or of their equal (Constr. and 34. 1) CB, and CK the square of CB, because BK the side of the square BN is equal to CB, and KO its equal (43. 1), also equal to the square of CB; the four rectangles CK, BN, GR, KO are each equal to the square of CB; again, AG, MP are rectangles under AC CB, because CG, GP are each equal to CB, being sides of the squares just ' mentioned, and MG to AC (34. 1), and the rectangle PL is equal to MP (43. 1), and therefore equal to the rectangle under AC, CB; also RF is equal to the rectangle under AC, CB, because RO is equal to BD (34. 1.), or CB, and RL is (by the same) equal to PH a side of the square XH, on XP equal to AC; therefore the four rectangles AG, MP, PL, RF are each equal to the rectangle AG under AC, CB; therefore the four squares CK, BN, GR, KO with the four rectangles AG, MP, PL, RF, or, which is equal, the gnomon AOй, is equal to four times the square CK and rectangle AG, or, which is equal (3. 2), four times the rectangle AK; whence, four times the rectangle AK being equal to the gnomon AOH, adding to each the square XH, four times the rectangle AK with the square