PROPOSITION. VIII. THEOREM. Triangle, from the Right Angle to the Base, 1 LETA BC be a Right-angled Triangle, whose Right Angle is BAC; and let the Perpendicular A D, be drawn from the Point A to the Base B C. I fay the Triangles ABD, ADC, are similar to one another, and to the whole Triangle ABC. For, because the Angle B A C is equal to the An. gle A D B, for each of them is a Right Angle ; and the Angle at B is common to the two Triangles A B C, A BD; the remaining Angle A C B shall be * equal to the remaining Angle B AD. Therefore * Cor. 32. 16 the Triangle A B C is equiangular to the Triangle ABD; and so, as + BC, which fubtends the Right + 4 of this. Angle of the Triangle A B C, is to B A, subtending the Right Angle of the Triangle ABD, fo is AB, fubtending the Angle C of the Triangle A BC, to DB, subtending an Angle equal to the Angle C, viz. the Angle B AD, of the Triangle ABD; and so, moreover, is AC to AD, subtending the Angle B, which is common to the two Triangles. Therefore the Triangle A B C is fequiangular to the Triangle 1 Def. in ABD; and the sides about the equal Angles are this. proportional. Wherefore the Triangle A B C is I also similar to the Triangle ABD. By the same Way we demonftrate, that the Triangle ADC is also fimilar to the Triangle A B C. Wherefore each of the Triangles A B D, A DC, is similar to the whole Triangle. I say, the said Triargles are also similar to one rother, For, because the Right Angle B D A is equal to the Right Angle ADC, and the Angle BAD has been proved equal to the Angle C; it follows, that the remaining Angle at B * Ihall be equal to the remaining * Cor. 32. 1. Angle Angle D AC. And so the Triangle A B D is equi† 4 of sbiso angular to the Triangle A DC. Wherefore as + BD, subtending the Angle B A D of the Triangle ABD, is to D A, subrending the Angle at C of the Triangle ADC, which is equal to the Angle B AD; fo is A D, fubtending the Angle B of the Triangle ABD, to D C, subtending the Angle D A C, equal to the Angle B. And, moreover, so is B A to AC, fubtending the Right Angles at D; and, consequently, the Triangle A B D is similar to the Triangle A DC. Wherefore, if a Perpendicular be drawn in a Rightangled 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 any 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 a Right Angle, is a mean Proportional between the whole Base, and the Segment thereof, which is next to the Side. PROPOSITION IX. PROBLEM Line. given, from which must be cut off any required Part; suppose a third. Draw any Right Line AC from the Point A, making an Angle at Pleasure with the Line A B. Affume any Part Din the Line A C; make * DE, EC, each # 31. 1. equal to AD; join BC, and draw + D F through D, parallel to B Č. Then, because F D is drawn Parallel to the Side | 2 of tbis, BC of the Triangle A B C, it Thall be, I as C D is to DA, fo is BF to F A. But CD is double to D A. Therefore B F shall be double to FA; and so B A is triple to AF. Wherefore, there is cut off A F, a third Part required of the given Right Line AB; which was to be done. PRO 3. I. PROPOSITION X. PROBLEM. other given Right Line is divided. A C a divided Line. It is required to divide A B, Let A C be divided in the Points D and E, and so placed, as to contain any Angle with A B. Join the Points C and B ; through D and E let DF, ÉG, be drawn * parallel to B Č; and thro' D draw D HK,* 31.1. parallel to A B. Then FH, HB, are each of them Parallelograms; and so D H is t equal to FG, and H K to G B. And +34. 1. because HE is drawn parallel to the Side KC of the Triangle D KC, it shall be I as CE is to E D, so is 1 2 of tbis. KH to HD. But K H is equal to B G, and H D to GF. Therefore, as CE is to ED, so is B G to GF. Again, because F D is drawn parallel to the Side E G of the Triangle A GE, as E D is to D A, fo fhall + GF be to F A. But it has been proved, that CE is to ED, as B G is to GF. Therefore, as CE is to E D, fo is B G to GF; and as E D is có DA, so is G F to FA. Wherefore, the given undivided Line A B is divided as the given Line A Cis; which was to be done. PROPOSITION XI. PROBLEM. Proportional to thim. placed, as to make any Angle with each other. Produce AB, AC, to the Points D and E, make Then, Then, because B C is drawn parallel to the Side | 2 of this. DE of che Triangle ADE, it fhall be, I as A'Bis to BD, so is AC to CE. But B D is equal to AC Hence, as A B is to AC, so is A C to CE. Therefore, a third Proportional CE is found to two given Right Lines A B, A C; which was to be done. PROPOSITION XII. PROBLEM. Proportional to ibem. required to find a fourth Proportional to them. PROPOSITION XIII. PROBLEM Right Lines. is required to find a mean Proportional between then. Place AB, BC, in a direct Line ; and on the Whole AC describe the Semicircle ADC, and I draw B D at Right Angles to AC from the Point 111. I. B; and let AD, D C, be joined. Then, because the Angle AD.C, in a Semicircle, +31. 3. is + a Right Angle ; and since the Perpendicular DB is drawn from the Right Angle to the Bale; therefore DB DB is f a mean Proportional between the Segments of 1 Cor. 8. of the Base A B, B C." Wherefore, a mean Proportional ibis. between the two given Lines A B, BC, is found; which was to be done. PROPOSITION XIV. THEOREM. one equal to one Angle of the other, bave the cal, are equal LET A B, B.C, be equal Parallelograms, having B B E, be in one strait Line ; then also I will the Sides (14. t. FB, B G, 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 A D is to B E, lo is G B to BF. For, let the Parallelogram F E be compleated. Then, because the Parallelograms A B is equal to the Parallelogram B C, and F E is some other Parallelogram ; it shall be, as A B is to F E, so is + B C +7.5. to FE; but as A B is to F E, so is IBD to BE; and as B C is to F E, so is G B to B F. Therefore 11 of sbit. as D B is to BE, so is G B to B F. Wherefore the Sides of the Parallelograms A B, B C, that are about the equal Angles, are reciprocally proportional. And, if the Sides that are about the equal Angles are reciprocally proportional, viz. if D B be to B E, as GB ir to BF; I say, the Parallelogram A B is equal to the Parallelogram BC. For, fince D B is to B E, as G B is to BF; and D B to B E, as the Parallelogram A B I to the Parallelogram F E, and G B I to B F, as the Parallelogram B C to the Parallelogram FE; it shall be as AE is to FE, fo is B C F to FE. Therefore the Parallelogram A B is equal to the Parallelogram B C. · And so, equal Parallelograms, having one Angle of the M 2. one |