PROP. V. PROB.

Fig. 6. 7. About a given triangle (BAC) to circumscribe a circle. & 8.

(1) Const

B. 1.

Bisect any two sides BA and AC of the given triangle, and through the points of bisection D and E draw DF and EF perpendicular to AB and AC, and from their point of concourse F draw to any angle A of the triangle BAC the line FA; the circle described from the centre F with the radius FA is circumscribed about the given triangle.

Draw FB and FC. In the triangles FDA, FDB the sides DA and DB are equal (1), FD is common to both and the angles at D are right (1,) therefore the (2) Prop. 4. sides FA and FB are equal (2): in the same manner it can be demonstrated that the lines FA and FC are equal, therefore the three lines FA, FB, and FC are (3) 4r. 1. equal (3), and therefore the circle described from the centre F with the radius FA passes through B and C, and therefore is circumscribed about the given triangle BAC.

Fig. 6.

Fig. 7.

Schol. This problem is the same as to describe a circle through three given points, which are not in one right line.

Cor. 1. If the centre F fall within the triangle, it is evident that all the angles are acute, for each of them is in a segment greater than a semicircle (1). If the centre F be in any side of the triangle, the angle op(1)Prop.31 posite to that side is right, because it is an angle in a semicircle (1). And if the centre fall without the triangle, the angle opposite to the side which is nearest to the centre is obtuse, because it is an angle in a segment greater than a semicircle (1).

B. 3.

Fig. 8.

Fig. 9.

Cor. 2. A circle can be circumscribed about a quadrilateral figure AFDC, whose opposite angles are equal to two right angles, for draw the diagonal and the circle described about the triangle FACmust pass through D. For take any point B in the arch FDC and draw BF and BC. In the quadrilateral AFBC inscribed in a circle the angles A and B are equal to two right angles (2) hypoth. (1), but D and A are equal to two right angles (2),

(1)Prop.22 B.3.

therefore the angles B and D are equal, and because the angle B is in the circumference, the angle D must also be in the same circumference (3), therefore the cir- (3) Cor. cle passes through D and therefore is circumscribed Prop. 21. about the given quadrilateral figure AFDC.

PROP. VI. PROB.

In a given circle (ABCD) to inscribe a square.

Draw any diameter AC of the given circle, draw BD perpendicular to it, and join AB, BC, CD, DA. ABCD is a square inscribed in the given circle.

B. 5.

Fig. 10.
See N.

(1) Schol.

Because the angles at E are right and therefore equal, the arches on which they stand are equal (1), and therefore their subtenses are equal (1); the figure ABCD is Prop. 29. therefore equilateral: and because BD is a diameter, B. 3. the angle BAD is in a semicircle and therefore right (2); in the same manner it can be demonstrated that the angles B, C and D are right, therefore, since the 3. sides are also equal, the figure ABCD is a square.

PROP. VII. PROB.

(2) Prop.31.

About a given circle (ABCD) to circumscribe a square. Fig.11.

Draw any diameter AC of the given circle and BD perpendicular to it, and through their extremities A, B, C, and D, draw the lines KF, FG, GH and HK tangents to the circle, the figure FGHK is a square circumscribed about the given circle.

See N.

Pop.

18. B. 3.

Because EA is drawn from the centre to the point of contact, the angle EAF is right (1), but the angle AEB is also right (2), therefore the lines FK and BD are (2) Constr parallel (3); in the same manner it can be demon- (3) Prop. strated that GH is parallel to BD, and also that FG 28. B. i. and KH are parallel to AC, therefore GD, BK, FC and AH are parallelograms, and because the angles at A are right, the angles at G and H opposite to them are right (4); in the same manner it can be demonstrated (4) Prop. that K and F are right angles, therefore FGHK is a

M

34 B. 1.

Fig. 11.

B. 1.

rectangle, and because AC and BD are equal, and FK and GH are equal to BD, and FG and KH are equal to AC (4), it is evident that FGHK is also equilateral, and therefore a square.

Cor. The square circumscribed is double of the square inscribed in the same circle, for it is equal to the square of the diameter of the circle, but the square of the diameter is double of the inscribed square, as is evident from prec. prop. and prop. 47. B. 1.

PROP. VIII. PROB.

In a given square (FGHK) to inscribe a circle.

Bisect two adjacent sides GH and FG of the given square in C and B, through C draw CA parallel to either FG or KH, and through B draw BD parallel to either GH or FK; the circle described from the centre E with the radius EC is inscribed in the given square.

Because GE, EH, EK and EF are parallelograms (1) Constr. (1), their opposite sides are equal (2), therefore CE and (2) Prop.34. EB are equal to GB and GC, but GB and GC are (3) Const. equal, for they are halves (3) of the equal lines FG and GH, therefore CE and EB are equal, but ED and EA are equal to CE and EB, for they are equal to CH and BF the halves of GH and FG (2), therefore the four lines EC, EB, EA and ED are equal, and therefore the circle described from the centre E with the radius EC passes through B, A and D, and because the angles at C, B, A and D are right, the sides of the (4)Prop. 16. square are tangents to the circle (4), which is therefore inscribed in the given square.

B 3.

Fig. 10.

PROP. IX. PROB.

About a given square (ABCD) to circumscribe a circle.

Draw AC and BD intersecting one another in E, the circle described from the centre E with the radius EA must pass through B, C and D.

For, since ABC is an isosceles triangle, and the angle B is right; the other angles are each half a right angle (1), in the same manner it can be demonstrated (1) Cor. 3. that each of the angles into which the angles of the Prop. 32. square are divided, is half a right angle; they are B. 1. therefore all equal; and therefore in the triangle AEB, as the angles EAB and EBA are equal, the sides EA and EB are equal (2), in the same manner it can be de- (2) Prop, 6. monstrated that ED and EC are equal to EA and EB, B.1. therefore the four lines EA, EB, EC and ED are equal, and therefore the circle described from the centre E with the radius EA passes through B, C and D, and is circumscribed about the given square.

PROP. X. PROB.

To construct an isosceles triangle, in which each of the Fig. 12. angles at the base shall be double of the vertical angle.

Take any line AB and divide it in C, so that the rectangle under AB and CB shall be equal to the square of AC (1), from the centre A with the radius (1)Prop.11.

AB describe a circle BED and inscribe in it a line BD B. 2. equal to AC (2), join AD, and BAD is an isosceles (2) Prop. 1. triangle in which the angles B and D are each double B. 4. of the angle A.

Draw DC and circumscribe a circle ACD about the triangle DCA.

Because the rectangle under AB and BC is equal to

the square of AC (3), or to the square of BD (3), the (3) Constr. line BD is a tangent to the circle ACD (4), and there- (4) Prop.37. fore the angle BDC is equal to the angle A in the al- B. 3. ternate segment (5), add to both the angle CDA, and (5) Prop.32. BDA is equal to the sum of the angles CDA and A, B. 3. but, since the sides AB and AD are equal, the angles B and BDA are equal, therefore the angle B is equal to the sum of CDA and A; but the external angle BCD is equal to the sum of CDA and A (6), therefore the (6) Prop.32. angles B and BCD are equal, and therefore the sides B. 1. BD and CD are equal (7), but BD and CA are equal (8), (7) Prop. 6. therefore CD and CA are equal, and therefore the angles (8) Constr. A and CDA are equal; but BDA is equal to the sum of

B. 1.

the angles A and CDA, therefore it is double of A, and therefore the angle B is also double of A.

Schol. It is evident that the triangle BDC is isosceles and that each of the angles at the base is double of the angle BDC at the vertex; also that the triangle ACD is isosceles, and that the angle at the vertex ACD is triple of each of the angles at the base, for it is equal to the (1) Prop.32, sum of the angles B and BDC (1), of which one is double of the angle A, the other equal to it.

B. 3.

Fig. 13. See N.

Cor. Hence is derived the method of constructing upon a given base BD an isosceles triangle, in which the angle at the base shall be double the angle at the vertex; construct a triangle of this kind CDB, in which the given line shall be a side; produce BC, and make the angle BDA equal to the angle B, BA and DA must meet, and the triangle required is constructed.

PROP. XI. PROB.

In a given circle (ABCDE) to inscribe an equilateral and equiangular pentagon.

Construct an isosceles triangle, in which each of the angles at the base shall be double of the angle at the (1)Prop. 10. vertex (1), and inscribe in the given circle a triangle equiangular to it ACE (2), bisect the angles at the base A, and E by the right lines AD and EB, and join AB, BC, CD, DE and EA.

B. 4.

(2) Prop. 2. B. 4.

(3) Constr.

(4) Schol.

Prop. 29.
B. 3.

(5) Schol. Prop 29.

B. 5.

Because each of the angles CAE and CEA is double of ECA (3), and is bisected, the five angles CEB, BEA, ACE, CAD and DAE are equal, and therefore the arches upon which they stand are equal (4), and therefore the lines CB, BA, AE, ED and DC which subtend these arches are equal (4), and therefore the pentagon ABCDE is equilateral.

And because the arches AB and DE are equal, if the arch BCD be added to both, the arch ABCD is equal to BCDE, and therefore the angles AED and BAE standing upon them are equal (5), in the same manner it can be demonstrated that all the other angles are equal, and therefore the pentagon is also equiangular.