PROP. II. PROB.

In a given circle (BAC), to inscribe a triangle, equiangular to a given triangle (EDF).

Draw the right line GH, touching the circle in any point A (17. 3), and at the point A, with the right line AH, make the angle HAC equal to the angle E (23. 1); and E at the same point, with the right line AG, make the angle GAB equal to angle F, and join BC.

The angle E is equal to the angle HAC (Constr.), or, (s 3), to the angle B in the alternate segment; for a like reaso the angles F and C are equal; therefore the remaining angle is equal to the remaining BAC (32. 1); therefore the triang BAC, which is inscribed in the given circle (Def. 3. 4) is eq angular to the given triangle EDF.

PROP. III. PROB.

About a given circle (ABC), to circumscribe a triangle, equian lar to a given triangle (EDF),

Prod e any side EF, of the given triangle, both way to G and H; find the centre K of the given circle (1. 3), which draw any radius KA, with which, at K, make the BKA equal to DEG (23. 1); and, with BK at K, the BKC equal to DFH; and draw ML, MN and LN, tou the circle in the points A, B and C (17. 3).

a semicircle, and therefore right [31. 3]; and, in the case of fig. 3, the angle BA, opposite BC, is in a segment less than a semicircle, and therefore obtuse [31. 3].

Cor. 2.-The rectangle under any two sides [AC, CB] of a triangle [ABC], is equal to the square of a right line [CD], bisecting the angle [ACB] included by them, drawn from that angle, to the opposite side [AB], together with the rectangle [ADB], under the segments of the side [AB], to which the right line is so drawn.

A

E

D

B

About the triangle ABC circumscribe a circle [5. 4], whose circumference let CD produced meet in E, and join EB.

The triangles ADC. EBC, having the angles ACD, ECB equal [Hyp.]. as also the angles CAD, CEB, being in the same segment CAEB [21. 3], are equiangular (32. 1); therefore, the rectangles under the sides about the equal angles ACD, ECB, taken alternately, are equal Cor. 4. 5 and 6. 2), namely, the rectangle under AC and CB, to the rectangle under CD and CE; but the rectangle under CD and CE, is equal to the square of CD with the rectangle CDE (3. 2); and the rectangle CDE is equal to the rectangle ADB (35. 3); therefore the rectangle under AC and CB, is equal to the square of CD, with the rectangle ADB.

Cor. 2.-The rectangle under any two sides (AC, CB, of a triangle (ABC), is equal to the rectangle under the perpendicular (CD), let fall from the angle (ACB), included by them, on the opposite side (AB), and the diameter of the circle, circumscribed about the triangle.

About the triangle ABC circumscribe a circle (5. 4), and, having drawn its diameter CE, join EB.

D

The triangles ACD, ECB, having the right angle ADC equal to the angle EBC in a semicircle 31. 3), and the angles CAD, CEB, in the same segment CAEB, equal (21. 3), are equiangular (32. 1); therefore the rectangles under the sides about the equal angles ACD, EC, taken alternately, are equal (Cor. 4. 5 and 6. 2), namely, the rectangle under AC and CB, to the rectangle under CD and CE.

Join BC. The angle CBA is equal to the angle EAK (32. 3), or its equal (29. 1) AED; therefore, the triangles ABC, AED, having besides, the angle BAE common, are equiangular; and therefore the rectangles under the sides about the, common angle BAE, taken alternately, are equal (Cor. 4. 5 and 6. 2), namely, the rectangle DAB to the rectangle EAC.

The demonstration is the same, if the right line parallel to the tangent HK, be without the circle, as fg; only substituting the small letters d, e, for their respective capitals, and, instead of the words, "the common angle BAE," using the words, "the equal vertical angles BAC, eAd."

Scholium.-If a right line be drawn from A, to a point L, in which FG meets the circle; it may, in like manner, be shewn, that the square of AL is equal to either of the rectangles DAB or EAC; for, drawing LB, the triangles ALD, ABL, having the angle LAB common, and the angle DLA equal to LAH (29. 1), or its equal (32. 3) LBA, are equiangular; and so the rectangles under the sides about the common angle LAB, taken alternately, are equal [Cor. 4. 5 and 6. 2], namely, the rectangle DAB to the square of LA.

PROP. VI. PROB.

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

Draw two diameters AC, BD of the given circle, at right angles to each other, and join AB, BC, CD, DA; ABCD is a square inscribed in the given circle.

For, since the triangles ABE, BCE, CDE, EDA, have their angles at the centre E equal, being right ones (Theor. at 11. 1), and also the sides containing them EA, EB, EC, ED (Def. 10. 1), the bases AB, BC, CD, DA are equal [4. 1]; therefore ABCD is equilateral, and the angles DAB, ABC, BCD, CDA are right, being angles in a semicircle (31. 3), there

B

fore ABCD is a square (Def. 36. 1), and inscribed in the given circle (Def. 3. 4).

Scholium.-In like manner, as in this proposition, the equality of the angles of the quadrangle ABCD, follows from the equality of the sides, it may be shewn, that any equilateral figure, inscribed in a circle, is also equiangular, and therefore [Def. 7. 4] regular; for each of the angles of the figure stand on an arch composed of the arches subtending all the sides of the figure, except two; which sides being equal [Hyp.], the arches subtending each of them are equal [28, 3], and therefore the whole arches on which the angles stand, and, of course, the angles themselves (27. 3).

PROP. VII. PROB.

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

Draw two diameters AC, BD of the given circle, at right angles to each K other, and through their extremes A, B, C, D, let tangents to the circle KF, FG, GH, HK be described [17. A 3]; FGHK is a square, circumscribed about the given circle.

T

For since EA is drawn from the centre to the contact, the angle EAF is right [18. 3], but the angle AEB is right [Constr.], therefore FK and BD are parallel [28. 1]; in like manner, GH may be proved parallel to BD, and FG and KH to AC, of course FD, BH, AH, FC are parallelograms [Def. 35. 1]; and, because the angles at A are right, the angles G and H opposite to them are right [34. 1]; in like manner, it may be proved, that the angles K and F are right, therefore the quadrangle FGHK is right angled; and since AC and BD are equal, and FG, KH are each equal to AC, and FK, GH each equal to BD [34. 1], the four sides FG, GH, HK, KF are equal to each other, and the quadrangle FGHK equilateral; it is therefore a square [Def. 36. 1], and circumscribed about the given circle [Def. 5. 4].

PROP. VIII. PROB.

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

Bisect two adjacent sides AB, AD in E and H; through E, draw EG parallel to AD or BC; and through H, HF parallel to DC or AB, meeting EG in K ; a circle described from the centre K, at the distance KE, is H inscribed in the given square.

For, since AK, KC, DK and KB are parallelograms [Constr.], the four right lines KE, KF, KĞ, KH

K

B