angular to the Triangle ADC. Wherefore as + BD + 4 of this. fubtending the Angle BAD of the Triangle ABD is to DA, fubtending the Angle at C of the Triangle ADC, which is equal to the Angle BAD, so is AD subtending the Angle B of the Triangle ABD to DC, fubtending the Angle DAC equal to the Angle B. And moreover, so is B A to AC, subtending the Right Angles at D; and consequently, the Triangle ABD is similar to the Triangle ADC. Wherefore, if a Perpendicular be drawn, in a Right-angled Triangle, from the Right Angle to the Base, then the Triangles on each side of the Perpendicular are similar both to the whole, and also to one another ; which was to be demonstrated. Coroll. From hence it is manifest, that the Perpendi cular drawn in a Right-angled Triangle from the Right Angle to the Base, is a mean Proportional between the Segments of the Base. Moreover, either of the Sides containing the Right Angle is a mean Proportional between the whole Base, and that Segment thereof which is next to the Side. PROPOSITION IX. PROBLEM. Line. ET AB be a Right Line given; from which must Draw any Right Line AC from the Point A, making an Angle at Pleasure with the Line AB. Ar fume any Part D in the Line AC, make* DE, EC, * 3. I. cach equal to AD, join BC, and draw + DF thro' t 31. 1. D, parallel to BC. Then because FD is drawn parallel to the Side BC of the Triangle ABC, it shall be $ as CD is 2 of tbis. to DA, fo is BF to F A. But CD is double to DA. Therefore BF shall be double to FA; and fo B A is triple to AF. Wherefore there is cut off AF, a third Part required of the given Right Line AB; which was to be done. PRO PROPOSITION X. PRO È LE M. other giveň Right Line is divided. ET A B be a given undivided Right Line, and AC a divided Line. It is required to divide AB, as AC is divided. Let A C be divided in the Points D and E, and so placed, as to contain any Angle with AB. Join the Points C and B; thro' D and E let DF, ÉG, be drawn * parallel to BC; and thro’ D, draw DHK, Then FH, HB, are each of them Parallelograms; # 34.1. and so DH is † equal to FG, and HK to GB. And because HE is drawn parallel to the Side KC, of the 12 of this. Triangle DKC, it shall be fas CE is to ED; fo is KH to HD. But KH is equal to BG, and HD to . PROBLEM. proportional to them. placed, as to make any Angle with each other. Produce AB, AC, to the Points D and E; make BD equal to A C, join the Points B, C, and draw the Right Line D E thro' D parallel to B C. Then Then becaụse BC is drawn parallel to the Side DE, of the Triangle ADE, it shall be * as AB is to * 2 of this. BD, fo is AC to CE. But BD is equal to AC. Hence as A B is to AC, fo is AC to CE. Therefore a third proportional CE is found to two giver Right Lines AB, AC; which was to be done. PROPOSITION XII. PROBLEM. proportional to them. LE ET A, B, C, be three Right Lines given. It is required to find a fourth proportional to them. Let DE and DF be two Right Lines, making any Angle EDF with each other. Now make DG equal to A, GE equal to B, DH equal to Ċ, and draw the Line GH, as also + EF thro' E, parallel to GH, tzn. t. Then because GH is drawn parallel to EF, the Side of the Triangle DEF, it shall be as D G is to GE, fo is DH to HF. But DG is equal to A, GE to B, and DH to C. Consequently as A is to B, fo is C to HF. Therefore the Right Line HF, X fourth proportional to the three given Right Lines A, B, C, is found; which was to be done. PROPOSITION XIII. PRO B L È M. Right Lines. ET the two given Right Lines be AB, BC. It them. Place AB, BC, in a direct Line, and on the whole AC describe the Semicircle ADC, and * draw * 11. Io BD at Right Angles to AC from the Point B, and let AD, DC, be joined. Then because the Angle ADC, in a Semicircle, is t a Right Angle, and since the Perpendicular † 350 g DB is drawn from the Right Angle to the Base ; therefore * Cor. 8. of therefore D B is * a mean Proportional between the this, Segments of the Base AB, BC. Wherefore a mean Proportional between the two given Lines AB, BC, is found; which was to be done. PROPOSITION XIV. * 14. 1. 47.50 THEOREM. Equal Parallelograms having one Angle of the one equal to one Angle of the other, bave the sides about the equal Angles reciprocal ; and those Parallelograms that have one Angle of the one equal to one Angle of the other, and the Sides that are about the equal Angles reciprocal, are equal between themselves. ET AB, BC, be equal Parallelograms, having the Angles at B equal; and let the Sides DB, BĚ, be in one strait Line; then also will * the Sides FB, BG, be in one strait Line. I say, the Sides of the Parallelograms AB, BC, that are about the equal Angles, are reciprocal; that is, as DB is to BE, fo is GB to BF. For let the Parallelogram FE be compleated. Then because the Parallelogram AB is equal to the Parallelogram BC, and FĒ is some other Paral lelogram; it shall be as AB is to F E, so is + BC to 1 of tbis. FE; but as AB is to FE, fo is IDB to BE; and as B C iš to F E, so is GB to BF. Therefore, as DB is to BE, so is GB to BF. Wherefore the Sides of the Parallelograms AB, BC, that are about the equal Angles, are reciprocally proportional. And if the sides that are about the equal Angles are reciprocally proportional, viz. if BD be to B E as GB is to BF: I say the Parallelogram AB is equal to the Parallelogram BC. For fince DB is to BE as GB is to BF, and DB to BE as the Parallelogram AB I to the Parallelogram FE, and GB I to BF as the Parallelogram BC to the Parallelogram FE; it shall be as AB is to FE, fo is BC to FE. Therefore the Parallelogram A B is equal to the Parallelogram B.C. And so equal Parallelograms having one Angle of the one equal to one Angle of the other, have the Sides about the equal An gles 2 gles reciprocal ; and those Parallelograms that have one Angle of the one equal to one Angle of the other, and the Sides that are about the equal Angle reciprocal, are equal betwern themselves; which was to be demonstrated. PROPOSITION XV. THE O R E M. Equal Triangles having one Angle of the one equal to one Angle of the other, have their Sides about the equal Angles reciprocal ; and those Triangles that have one Angle of the one equal tą one Angle of the other, and have also the Sides about the equal Angles reciprocal, are equal between themselves. Angle of the one equal to one Angle of the other, viz. the Angle BAC equal to the Angle DAE. I say the Sides about the equal Angles are reciprocal, that is, as CA is to AD, fo is EA to AB. For plače CA and AD in one strait Line, then EA and A B shall be * also in one strait Line, and let * 14. 1. BD be joined. Then because the Triangle ABC is equal to the Triangle ADE, and ABD is some other Triangle, the Triangle CAB shall be † to the Tri- +7.50 angle B AD, as the Triangle ADE is to the Triangle B AD. But as the Triangle CAB is to the Triangle BAD, so is CA Ito AD, and as the Tri- II of this. angle EAD is to the Triangle BAD, fo is I EA to AB. Therefore as CA is to AD, so is EA to AB. Wherefore the sides of the Triangles ABC, ADE, about the equal Angles, are reciprocal. And if the sides about the equal Angles of the Triangles ABC, ADE, be reciprocal, viz. if. CA be to AD as E A is to AB, I say the Triangle ABC is equal to the Triangle ADE. For, again let BD be joined. Then becaufe CA is to AD as E A is to AB, and CA to AD | as the Triangle ABC to the Triangle BAD, and 'E A to AB | as the Triangle EAD to the Triangle BAD; therefore, as the Triangle ABC is to the Triangle BAD, fo Ahall the Triangle EAD be to the Triangle M BAD |