ing angle AMB is equal to the remaining angle DEF. In like manner, the angle LMN may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal (2. Cor. 25. 1.) to the remaining angle EDF: Wherefore the triangle LMN is equiangular to the triangle DEF: And it is described about the circle ABC.

PROP. IV. PROB.

To inscribe a circle in a given triangle.

Let the given triangle be ABC; it is required to inscribe a circle in ABC.

A

E

D

Bisect (Prob. 4. 1.) the angles ABC, BCA by the straight lines BD, CD meeting one another in the point D, from which draw (Prob. 7. 1.) DE, DF, DG perpendiculars to AB, BC, CA. Then because the angle EBD is equal to the angle FBD, the angle ABC being bisected by BD; and because the right angle BED, is equal to the right angle BFD, the two triangles EBD, FBD have two angles of the one equal to two angles of the other; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides are equal; wherefore DE is equal to DF. For the same reason, DG is equal to DF; therefore the three straight lines DE, DF, DG, are equal to one another, and the circle described from the centre D, at the distance of any of them, will pass through the extremities of the other two, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G, are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches (1. Cor. 16. 3.) the circle. Therefore the straight lines AB, BC, CA, do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC.

SCHOLIUM.

B

Lines bisecting the three angles of a triangle, all meet in the same point; that is, the centre of the inscribed circle and it has been shown (Sch. Prob. 7. B. 3.) that the lines bisecting the three sides at right angles, also meet in one point; consequently, in the equilateral triangle, since the lines bisecting the angles also bisect the sides, it follows that the centre of the circumscribed and inscribed circles coincide.

PROP. V. PROB.

To inscribe a square in a given circle.

Let ABCD be the given circle; it is required to inscribe a square in ABCD.

E

D

Draw the diameters AC, BD at right angles to one another, and join AB, BC, CD, DA; because BE is equal to ED, E being the centre, and because EA is at right angles to BD, and common to the triangles ABE, ADE; the base BA is equal (1. 1.) to the base AD; and, for the same reason, BC, CD are each of them equal to BA or AD; therefore the quadrilateral figure ABCD is equilateral. It B is also rectangular; for the straight line BD being a diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right (29. 3.) angle; for the same reason each of the angles ABC, BCD, CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular, and it has been shewn to be equilateral; therefore it is a square; and it is inscribed in the circle ABCD.

C

SCHOLIUM.

Since the triangle AED is right angled and isosceles, we have (Cor. 2. Th. 37. B. 1.) AD: AE:: √2:1; hence the side of the inscribed square is to the radius, as the square root of 2, is to unity.

PROP. VI. PROB.

To describe a square about a given circle.

Let ABCD be the given circle; it is required to describe a square

about it.

Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D draw (Prob. 2. 3.) FG, GH, HK, KF touching the circle; and because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contact A, the angles at A are right (18. 3.) angles; for the same reason, the angles at the points B, C, D are right angles; and because the angle AEB is a right angle, as likewise is EBG, GH is parallel (20. 1.) to

B

F

D

D

AC; for the same reason, AC is parallel to FK, and in like manner, GF, HK may each of them be demonstrated to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms; and GF is therefore equal (28. 1.) to HK, and GH to FK; and because AC is equal to BD, and also to each of the two GH, F'K; and BD to each of the two GF, HK: GH, FK are each of them equal to GF or HK; therefore the quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB (28. 1.) is likewise a right angle: In the same manner, it may be shewn that the angles at H, K, F are right angles; therefore the quadrilateral figure FGHK

H

K

is rectangular; and it was demonstrated to be equilateral; therefore it is a square; and it is described about the circle ABCD.

PROP. VII. PROB.

To inscribe a circle in a given square.

Let ABCD be the given square; it is required to inscribe a circle in ABCD.

E

D

Bisect (Prob. 5. 1.) each of the sides AB, AD, in the points F, E, and through E draw (Prob. 13. 1.) EH parallel to AB or DC, and through F draw FK parallel to AD or BC; therefore each of the figures, AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are equal (28. 1.); and because that AD is equal to AB, and that AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the sides opposite to these are equal, viz. FG to GE; in the same manner it may be demonstrated, that GH, GK, are each of them equal to FG or GE; therefore the four straight lines, GE, GF, GH, GK, are equal to one another; and the circle described from the centre G, at the distance of one of them, will pass through the extremities of the other three; F and will also touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K, are right (21. 1.) angles, and because the straight line which is drawn from the extremity of a diameter at right angles to it, touches the circle (16. 3.); therefore each of the straight lines AB, BC, CD, DA touches the circle, which is therefore inscribed in the squares ABCD.

B

H

C

PROP. VIII. PROB.

To describe a circle about a given square.

Let ABCD be the given square; it is required to describe a circle about it.

Join AC, BD, cutting one another in E; and because DA is equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC are equal to the two BA, AC, and the base DC is equal to the base BC; wherefore the angle DAC is equal to the angle BAC, and the angle DAB is bisected by the straight line AC. In the same manner it may be demonstrated, that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC; therefore, be

E

B

cause the angle DAB is equal to the angle ABC, and the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal

to the angle EBA: and the side EA (4. 1.) to the side EB. In the same manner, it may be demonstrated, that the straight lines EC, ED are each of them equal to EA, or EB; therefore the four straight lines EA, EB, EC, ED, are equal to one another; and the circle described from the centre E, at the distance of one of them, must pass through the extremities of the other three, and be described about the square ABCD.

PROP. IX. PROB.

To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part.

Let AB be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

F

G

Upon AB describe (Prob. 18. 1.) the square ABDC; bisect (Prob. 5. 1.) AC in E, and join BE; producs CA to F, and make (Prob. 3. 1.) EF equal to EB, and upon AF describe (Prob. 18. 1.) the square FGHA, AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. Produce GH to K: Because the straight line AC is bisected in E, and produced to the point F, the rectangle CF.FA, together with the square of AE, is equal (6. 2.) to the square of EF: But EF is equal to EB; therefore the rectangle CF.FA, together with the square of AE, is equal to the square of EB; And the squares of BA, AE are equal (37. 1.) to the square of EB, because the angle EÁB is a right angle; therefore the rectangle CF.FA, together with the square of AE, is equal to the squares of BA, AE: take away the square of AE, which is common to both, therefore the remaining rectangle CF.FA is equal to the square of AB. Now the figure E FK is the rectangle CF.FA, for AF is equal to FG; and AD is the square of AB; therefore FK is equal to AD: take away the common part AK, and the remainder FH is equal to the remainder HD. But HD is the rectangle AB.BH, for AB is equal to BD; and FH is the square of AH; therefore the rectangle AB.BH is equal to the square of AH: Wherefore the straight line AB is divided in H, so that the rectangle AB.BH is equal to the square of AH.

C

K

D

PROP. X. PROB.

To describe an isosceles triangle, having each of the angles at the base double of the third angle.

Take any straight line AB, and divide (9. 3.) it in the point C, so that the rectangle AB.BC may be equal to the square of AC; and from the centre A, at the distance AB, describe the circle BDE, in which place (1. 4.) the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; join DA, DC, and about the triangle ADC describe (Prob. 7. 3.) the circle ACD; the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double of the angle BAD.

B

A

Because the rectangle AB.BC is equal to the square of AC, and AC equal to BD, the rectangle AB.BC is equal to the square of BD; and because from the point B without the circle ACD two straight lines BCA, BD are drawn to the circumferences, one of which cuts, and the other meets the circle, and the rectangle AB.BC contained by the whole of the cutting line, and the part of it without the circle, is equal to the square of BD, which meets it; the straight line BD touches (33. 3.) the circle ACD. And because BD touches the circle, and DC is drawn from the point of contact D, the angle BDC is equal (30. 3.) to the angle DAC in the alternate segment of the circle, to each of these add the angle CDA; therefore the whole angle BDA is equal to the two angles CDA, DAC; but the exterior angle BCD is equal (25. 1.) to the angles CDA, DAC; therefore also BDA is equal to BCD; but BDA is equal to CBD, because the side AD is equal to the side AB; therefore CBD, or DBA is equal to BCD; and consequently the three angles BDA, DBA, BCD, are equal to one another. And because the angle DBC is equal to the angle BCD, the side BD is equal (4. 1.) to the side DC; but BD was made equal to CA; therefore also CA is equal to CD, and the angle CDA equal to the angle DAC; therefore the angles CDA, DAC together, are double of the angle DAC; but BCD is equal to the angles CDA, DAC (25. 1.); therefore also BCD is double of DAC. But BCD is equal to each of the angles BDA, DBA, and therefore each of the angles BDA, DBA, is double of the angle DAB; wherefore an isosceles triangle ABD is described, having each of the angles at the base double of the third angle.

"COR. 1. The angle BAD is the fifth part of two right angles. For "since each of the angles ABD and ADB is equal to twice the angle "BAD, they are together equal to four times BAD, and therefore all the "three angles ABD, ADB, BAD, taken together, are equal to five times "the angle BAD. But the three angles ABD, ADB, BAD are equal to