A C, and C E, together, fince A CE is a Right Angle. Therefore the Square of E A is double to the Square of A C. Again, becaufe E Gis equal to GF, and the Square of E G is equal to the Square of GF; therefore the Squares of EG, GF, together, are double to the Square of G F. But the Square of E F is equal to the Squares of EG, GF. Thereforet 47. 1 the Square of E F is double the Square of G F But GF is equal to CD; and fo the Square of EF is double to the Square of C D. But the Square of A E is likewife double to the Square of A C. Wherefore the Squares of A E, and E F, are double to the Squares of A C, and CD. But the Square of A F is + equal to the Squares of A E, and E F; because the Angle AEF is a Right Angle, and confequently the Square of A F is double to the Squares of A C, and CD. But the Squares of A D, and D F, are equal to the Square of AF: For the Angle at D is a Right Angle. Therefore the Squares of A D, and D F, together, fhall be double to the Squares of A C, and CD, together. But DF is equal to D B. Therefore the Squares of A D, and D B, together, will be double to the Squares of A C, and C D, together. Wherefore, if a Right Line be any bow cut into two equal, and two unequal Parts; then the Squares of the unequal Parts together, are double to the Squares of the half Line, and the Squares of the intermediate Part; which was to be demonftrated. PROPOSITION X. THEOREM. If a Right Line be cut into two equal Parts, and to it be directly added another; the Square made on the Line compounded of] the whole Line, and the added one, together with the Square of the added Line, fhall be double to the Square of the half Line, and the Square of [that Line which is compounded of] the balf, and the added Line. LE ET the Right Line A B be bifected in C, and any ftrait Line BD added directly thereto. I fay, the Squares of A D, and D B, together, are double to the Squares of AC, and CD, together. For, $ 49. 1. 12. I. EB. But the Squares of B A and A E are equal to the Square of EB; for the Angle at A is a Right Angle. Therefore the Rectangle under CF and F A, together with the Square of AE, is equal to the Squares of BA, and A E. And, taking away the Square of A E, which is common, the remaining Rectangle under C F and F A is equal to the Square of A B. But FK is the Rectangle under C F and FA; fince A F is equal to FG; and the Square of A B is AD. Wherefore the Rectangle FK is equal to the Square A D. And if A K, which is common, be taken from both, then the remaining Square F H is equal to the remaining Rectangle HD. But HD is the Rectangle under A B and B H, fince A B is equal to BD; and F H is the Square of AH. Therefore, the Rectangle under AB and BH fhall be equal to the' Square of A H. And fo the given Right Line A B is cut in H, fo that the Rectangle under A B and B H is equal to the Square of AH; which was to be done. PROPOSITION XII. THEOREM. In an obtufe-angled Triangle, the Square of the Side fubtending the obtufe Angle is greater than the Squares of the Sides containing the obtufe Angle, by twice the Rectangle under one of the Sides, containing the obtufe Angle, viz. that on which, produced, the Perpendicular falls, and the Line taken without, between the Perpen-. dicular and the obtufe Angle. LET ABC be an obtufe-angled Triangle, having the obtufe Angle BAC; and * from the Point B draw BD perpendicular to the Side C A produced. I fay, the Square of BC is greater than the Squares of BA, and A C, by twice the Rectangle contained under CA, and A D. For, because the Right Line CD is any how cut in +4 of this. the Point A, the Square of CD fhall be + equal to the Squares of CA, and A D, together with twice the Rectangle under C A, and A D. And if the Square of BD, which is common, be added, then the Squares of * of CD and DB are equal to the Squares of C A, AD, and DB, and twice the Rectangle contained under CA and A D. But the Square of CB is equal to 47. 8. the Squares of CD, DB; for the Angle at D is a Right one, fince BD is perpendicular; and the Square of A B is equal to the Squares of AD and D B. Therefore the Square of C B is equal to the Squares of CA, and A B, together with twice a Rectangle under C A, and AD. Therefore, in an obtufe-angled Triangle, the Square of the Side fubtending the obtufe Angle is greater than the Squares of the Sides containing the abtufe Angle, by twice the Rectangle under one of the Sides containing the obtufe Angle, viz. that on which, produced, the Perpendicular falls, and the Line taken without, between the Perpendicular and the obtufe Angle; which was to be demonftrated. PROPOSITION XIII. THEOREM. In an acute-angled Triangle, the Square of the Side fubtending the acute Angle is less than the Squares of the Sides containing the acute Angle, by twice a Rectangle under one of the Sides about the acute Angle, viz. that on which the Perpendicular falls, and the Line affumed within the Triangle, from the Perpendicular to the acute Angle. LET ABC be an acute-angled Triangle, having the acute Angle B; and from A let there be 12. 1. drawn AD perpendicular to B C. I fay, the Square of AC is less than the Squares of CB, and BA, by twice a Rectangle under C B, and B D. For, because the Right Line C B is cut any how in D, the Squares of C B, and B D will be † equal tot 7 of this. twice a Rectangle under C B, and BD, together with the Square of DC. And if the Square of AD be added to both, then the Squares of CB, BD, and DA, are equal to twice the Rectangle contained under CB, and BD, together with the Squares of AD, and D C. But the Square of AB is equal to the ₺ 47. Squares of B D, and D A; for the Angle at D is a 147. 1. Right Angle. And the Square of AC is equal to the Squares of AD and DC. Therefore the Squares of C B and B A are equal to the Square of A C, together with twice the Rectangle contained under C B and BD. Wherefore the Square of A C, only, is less than the Squares of CB and B A, by twice the Rectangle under CB and B D. Therefore, in an acuteangled Triangle, the Square of the Side fubtending the acute Angle is less than the Squares of the Sides containing the acute Angle, by twice a Rectangle under one of the Sides about the acute Angle, viz. that on which the Perpendicular falls, and the Line affumed within the Triangle, from the Perpendicular to the acute Angle; which was to be demonftrated. PROPOSITION XIV. PROBLEM. To make a Square equal to a given Right-lined LET A be the given Right-lined Figure. It is re- Make the Right-angled Parallelogram BCDE equal to the Right-lined Figure A. Now if B E be equal to ED, what was propofed will be already done, fince the Square B D is made equal to the Rightlined Figure A: But if it be not, let. either BE or ED be the greater: Suppofe BE, which let be produced to F; fo that E F be equal to E D. This be10. 1.ing done, let BF be + bifected in G, about which, as a Centre, with the Distance G B, or G F, defcribe the Semicircle BHF; and let DE be produced to H, and draw GH. Now, becaufe the Right Line B F is divided into two equal Parts in G, and into two unequal ones in E, the Rectangle under B E and E F, 15 of this. together with the Square of GE, fhall be equal to the Square of G F. But G F is equal to GH. Therefore the Rectangle under B E and E F, together, with the Square of G E, is equal to the Square of G H. But 47. 1. the Squares of H E and G E are equal to the Square of G H. Wherefore the Rectangle under B E and * EF, |