B : B to AB, AC; DF, EF produced meet one another : For, if they do not meet, they are parallel, wherefore AB, AC, which are at right angles to them, are parallel ; which is absurd : Let them meet in F, and join FA; also, if the point F be not in BC, join BF, CF: Then, because AD is equal to DB and DF common, and at right angles to AB, the base AF is equal (4. 1.) to the base FB: In like manner, it may be shewn that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, FC are equal to one another ; wherefore the circle described from the centre F, at the distance of one of them, shall pass through the extremities of the other two, and be described about the triangle ABČ. Which was to be done. Cor. And it is manifest, that when which it is, being in a segment less the centre of the circle falls within than a semicircle, is greater than a the triangle, each of its angles is less right angle: Wherefore, if the given than a right angle, each of them be- triangle be acute angled, the centre ing in a segment greater than a se- of the circle falls within it; if it be a micircle; but, when the centre is in right angled triangle, the centre is in one of the sides of the triangle, the the side opposite to the right angle; angle opposite to this side, being in a and, if it be an obtuse angled triangle, semicircle, is a right angle ; and, if the centre falls without the triangle, the centre falls without the triangle, beyond the side opposite to the obthe angle opposite to the side beyond tuse angle. PROP. VI. PROB. To inscribe a square in a given circle. Let ABCD be the given circle ; it ing the diameter of the circle ABCD, required to inscribe a square in BAD is a semicircle ; wherefore the ABCD. angle BAD is a right (31. 3.) angle; Draw the diameters AC, BD at for the same reason each of the right angles to one another; and join angles ABC, BCD, CDA is a right AB, BC, CD, DA; because BĚ is angle ; therefore the quadrilateral fiequal to ED, for E is the centre, and gure ABCD is rectangular, and it has that EA is common, and at right been shewn to be equilateral ; thereangles to BD; the base BA is equal fore it is a square ; and it is inscrib(4. 1.) to the base AD; and, for the ed in the circle ABCD. Which wss same reason, BC, CD are each of to be done. them equal to BA, or AD; therefore the quadrilateral figure ABCD is equi. lateral. It is also rectangular'; for the straight line BD, be PROP. VII. PROB. B D K с To describe a square about a given circle. Let ABCD be the given circle; it HK, and GH to FK; and because is required to describe a square about AC is equal to BD, it. and that AC is equal Draw two diameters AC, BD of to each of the two the circle ABCD, at right angles to GH, FK: and BD one another, and through the points to each of the two A, B, C, D, draw (17. 3.) FG, GH, GF, AK; GH, FK HK, KF touching the circle ; and are each of them ebecause FG touches the circle ABCD, qual to GF or HK; therefore the and EA is drawn from the centre E quadrilateral figure FGHK is equilato the point of contact A, the angles teral. It is also rectangular; for at A are right (18. 3.) angles ; for GBEA being a parallelogram, and the same reason, the, angles at the AEB a right angle, AGB (34. 1.) is points B, C, D are right angles; and likewise a right angle: In the same because the angle AEB is a right manner, it may be shewn that the angle, as likewise is EBG, GH is pa. angles at H, K, F are right angles ; rallel (28. 1.) to AC; for the same therefore the quadrilateral figure reason, AC is parallel to FK, and in FGHK is rectangular, and it was delike manner ĠF, HK may each of monstrated to be equilateral; therethem be demonstrated to be parallel fore it is a square; and it is described to BED; therefore the figures GK, about the circle ABCD. Which was GC, AK, FB, BK are parallelograms; done. and GF is therefore equal (34. 1.) to PROP. VIII. PROB. B H To inscribe a circle in a given square. Let ABCD be the given square ; GF, GH, GK are equal to one anoit is required to inscribe a circle in ther; and the circle described from the ABCD. centre G, at the disBisect (10. 1.) each of the sides tance of one of them, AB, AD, in the points F, E, and shall pass through the through E draw (31. 1.) EH parallel extremities of the oto AB or DC, and through F draw ther three, and touch FK parallel to AD or BC; therefore the straight lines AB, each of the figures AK, KB, AH, HD, BC, CD, DA; beAG, GC, BG, GD, is a parallelo- cause the angles at the points E, F, gram, and their opposite sides are H, K, are right (29. 1.) angles, and equal; (34. 1.) and because AD is that the straight line which is drawn equal to AB, and that AE is the half from the extremity of a diameter, at of AD, and AF the half of AB, AE is right angles to it, touches the.circle; equal to AF; wherefore the sides (18. 3.) therefore each of the straigbt opposite to these are equal, viz. FG lines AB, BC, CD, DA touches the to GE ; in the same manner, it may circle, which therefore is inscribed in be demonstrated that GH, GK are the square ABCD. Which was to each of them equal to FG or GE; be done. therefore the four straight lines GE, PROP. IX. PROB. To describe a circle about a given square. E Let ABCD be the given square; it verally bisected by the straight lines is required to describe a circle about BD, AC; therefore, because the angle it. DAB is equal to the angle ABC, and Join AC, BD cutting one another that the angle EAB is the half of in E; and because DA is equal to DAB, and EBA the half of ABC; the AB, and AC common to the triangles angle EAB is equal to the angle DAC, BAC, the two sides DA, AC EBA; wherefore the side EA is equal are equal to the two BA, AC, and (6. 1.) to the side EB: In the same the base DC is manner, it may be demonstrated, that equal to the base A D the straight lines EC, ED are each of BC; wherefore them equal to EA or EB; therefore the angle DAC the four straight lines EA, EB, EC, is equal (8.1.) to ED are equal to one another : and the angle BAC, the circle described from the centre E, and the angle B at the distance of one of them, shall DAB is bisected pass through the extremities of the by the straight line AC: In the same other three, and be described about mamer, it may be demonstrated that the square ABCD. Which was to be the ar gles ABC, BCD, CDA are se- done. PROP. X. PROB. To describe an isoceles triangle, having each of the angles at the buse double of the third angle. Take any straight line AB, and di- angle AB, BC, vide (11. 2.) it in the point C, so contained by the that the rectangle AB, BC, be equal whole of the cutto the square of CA; and from the ting line, and the centre A, at the distance AB, describe part of it without the circle BDE, in which place (1. 4.) the circle, is equal the straight line BD enual to AC, to the square of which is not greater than the diame- BD which meets ter of the circle BDE ; join DA, DC, it; the straight and about the triangle ADC describe line BD touches (37. 3.) the circle (5. 4.) the circle ACD; the triangle ACD; and because BD touches the ABD is such as is required, that is, circle, and DC is drawn from the each of the angles ABD, ADB is point of contact D, the angle BDC is double of the angle BAD. equal (32. 3.) to the angle DAC in Because the rectangle AB, BC is the alternate segment of the circle ; equal io the square of AC, and that to each of these add the angle CDA; AC is equal to BD, the rectangle AB, therefore the whole angle BDA is BC is equal to the square of BD; and equal to the two angles CDA, DAC; because from the point B without the but the exterior angle BCD is equal circle ACD, two straight lines BCA, (32. 1.) to the angles CDA, DÅC; BD are drawn to the circumference, therefore also BDA is equal to BCD; one of which cuts, and the other but BDA is equal (5. 1.) to the angle meets the circle, and that the reci- CBD, because the side AD is equal B to the side AB; therefore CBD, or are double of the angle DAC: But DBA is equal to BCD; and conse- BCD is tual to the angles CDA, quently the three angles BDA, DBA, DAC; therefore also BCĎ is double BCD, are equal to one another; and of DAC, and BCD is equal to each because the angle DBC is equal to of the angles BDA, DBA; each, the angle BCD, the side BD is equal therefore, of the angles BDA, DBA (6. 1.) to the side DC; but BD was is double of the angle DAB; wheremade equal to CA; therefore also CA fore an isosceles triangle ABD is deis equal to CD, and the angle CDA scribed, having each of the angles at egual (6. 1.) to the angle DAČ; there. the base double of the third angle. fore the angles CDA, DAC together, Which was to be done. PROP. XI, PROB. To inscribe an equilateral and equiangular pentagon in a given cirle. Let ABCDE be the given circle ; sected by the straight lines CE, DB it is required to inscribe an equilateral the five angles DAC, ACE, ECD and equiangular pentagon in the circle CDB, BDA are equal to one another, ABCDE. but equal angles stand upon equal Describe (10. 4.) an isosceles tri- (26. 3.) circumferences; therefore the angle FGH, having each of the angles five circumferences AB, BC, CD, DE, at G, H, double of the angle at F; EA are equal to one another; and and in the circle ABCDE inscribe equal circumferences are subtended (2. 4.) the triangle ACD equiangular by equal (29. 3.) straight lines ; thereto the triangle FGH, 80 that the anglefore the five straight lines AB, BC, CAD be equal to the angle at F, and CD, DE, EA are equal to one ano ther. Wherefore the pentagon ABCDE is equilateral. It is also equiauguılar; because the circumference AB is equal to the circumference DE: If to each be added BCD, ihe whole ABCD is equal to the whole EDCB; And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; thereeach of the angles ACD, CDA equal fure the angle SAE is equal (27. 3.) to the angle at G or H; wherefore to the angle AED: For the same reaeach of the angles ACD, CDA is son, each of the angles ABC, BCD, double of the angle CAD. Bisect CDE is equal to the angle BAE, or (9. ..) the angles ACD, CDA by the AED: Therefore the pentagon ABCDE straight lines CE, DB: and join AB, is equiangular; and it has been shewn BC, DE, EA. ABCDE is the pen. that'it is equilateral. Wherefore, in tagon required. the given circle, an equilateral and Because each of the angles ACD, equiangular pentagon has been inCDA is double of CAD, and are bi- scribed. Which was to be done, PROP. XII. PROB. To describe an equilateral and equiangular pentagon about a given circle. Let ABCDE be the given circle ; ral and equiangular pentagon about it is required to describe an equilate the circle ABCDE. Let the angles of a pentagon, in- ence BC is equal to the circumfer. scribed in the circle, by the last pro- ence CD, th:e angle BFC is equal position, be in the points A, B, C, D, (27. 3.) to the angle CFD: and BFC E, so that the circumferences AB, is double of the angle KFC, and BC, CD, DE, EA, are equal; (11. 4.) CFD double of CFL; therefore the and through the points A, B, C, D, E, angle KFC is equal to the angle draw GH, HK, KL, LM, MG, touch- CFL; and the right angle FČK ing (17. 3.) the circle ; take the centre is equal to the right angle FCL: F, and join FB, FK, FC, FL, D: Therefore, in the two triangles FKC, And because the straight line KL FLC, there are two angles of one touches the circle ABCDE in the equal to two angles of the other, each point C, to which FC is drawu from to each, and the side FC, which is the centre F, FC is perpendicular adjacent to the equal angles in each (18. 3.) to KL; therefore each of is common to both; therefore the the angles at C is a right angle: other sides shall be equal (26. 1.) to the For the same reason, the angles other sides, and the third angle to the at the points B, D are right angles; third angle: Therefore the straight and because FCK is a right angle, line KC is equal to CL, and the angle the square of FK is equal (47. 1.) FKC to the angle FLC : And because to the squares of FC, CK: For KC is equal to CL, KL is double of the same reason, the square of FK KC: In the same manner, it may be is equal to the squares of FB, BK: shewn that HK is double of BK: And Therefore the squares of FC, CK are because BK is equal to KC, as was equal to the squares of FB, BK, of demonstrated, and that KL is double which the square of FC is equal to of KC, and HK double of BK, HK the square of FB; the remaining shall be equal to KL: In like manner square of CK is therefore equal to the it may be shewn that GH, GM, ML remaining' square of BK, and the are each of them equal to HK or KL: straight line CK equal to BK: And Therefore the pentagon GHKLM is because FB is equal to FC, and FK equilateral. It is also equiangular; common to the triangles BFK, CFK, for, since the angle FKC is equal to the two BF, FK are equal to the two the angle FLC, and that the angle CF, FK; and the base BK is equal HKL is double of the angle FKC, to the base KC; therefore the angle and KLM double of FLC, as was BFK is equal (8. 1.) to the angle before demonstrated, the angle HKL KFC, and the angle BKF to FKC; is equal to KLM: And in like manwherefore the ner it may be shewn that each of angle BFC is the angles KHG, HGM, GML is double of the equal to the angle HKL or KLM: angle KFC, & Therefore the five angles GHK, HKL, BKC double of KLM, LMG, MGÅ being to FKC: For the one another, the pentagon GHKLM same reason, is equiangular : And it is equilateral, the angle CFD as was demonstrated ; and it is de is double of scribed about the circle ABCDE. the angle CFL, and CLD double of Which was to be done. CLF: And because the circumfer G E H M B D K PROP. XIII. PROB. To inscribe a circle in a given equilateral and equiangular pentagon. Let ABCDE be the given equila- required to inscribe a circle in the teral and equiangular pentagon ; it is pentagon ABCDE. |