terminated in A and D, by AH and DC; and draw KA, KB, KH, and KC, KG and KD: And because ABCD, AHGD, are quadrilateral figures described about the circle,.. (S. 20. 3.) the AKD+2BKC=two right angles; and, the AKD+ 2 HKG = two right angles; .. the BKC HKG; i. e. the subtended at the

L

centre by the tangent BC is equal to the tended at the centre by the tangent HG.

sub

30. COR. The two segments, which any two tangents, so drawn, cut off from the two given tangents, also subtend equal angles, at the centre of the circle.

Let BH and GC be the segments cut off from the tangents AH and DC, by the two tangents BC and GH: They subtend equal BKH, GKC at the centre K.

L

For it has been shewn that the BKC= 2 HKG; from these equals take away the common and there remains the BKH= 4 GKC.

PROP. XXII.

HKC,

31. PROBLEM. To draw a tangent to a given circle, such that its segment, contained between the point of contact, and an indefinite straight line, given in position, shall be equal to a given finite straight line.

Let ABC be a given circle, L a given finite straight line, and XY an indefinite straight line

given in position: It is required to draw a tangent to ABC so that its segment between the point of contact and XY shall be equal to L.

Find the centre K (E. 1. 3.) of the given circle, and take any diameter of it, as AKD; in AD produced find (S. 16. 3.) a point E from which if a tangent be drawn to the given circle ABC it shall be equal to L; from K as a centre, at the distance KE, describe the circle EFG, and let it meet, or cut, XY in F; from F draw (E. 17. 3.) FC to touch the circle ABC in C: And since (S. 19. 3.) FC is equal to the tangent which can be drawn from E, and which (constr.) is itself equal to L, it is manifest that CFL; i. e. the segment of the tangent between the point of contact C and XY is equal to the given straight line L.

PROP. XXIII.

32. THEOREM. If a straight line touch the interior of two concentric circles, and be terminated both

ways by the circumference of the outer circle, it shall be bisected in the point of contact.

Let GBC, EDH be two circles having a com

mon centre K, and let BC touch the interior circle EDH in D: Then is BC bisected in D.

For join K, D: And, because BC touches EDH in D, the KDC, KDB (E. 18. 3.) are right ; ‚. (E. 3. 3.) BC is bisected in D,

PROP. XXIV.

33. THEOREM. If a polygon be described about a circle, the straight lines joining the several points of contact will contain a polygon of the same number of angles as the former; and any two adjacent angles of the circumscribed figure shall be, together, the double of that angle, of the inscribed figure, which lies between them.

circle CLMED, in the several points C, L, M, E and D, and let these points be joined: Then it is manifest, that the polygon DCLME has the same number of angles as AFGHB; and, further, any two adjacent A and B of the polygon AFGHB, are, together, the double of the intermediate CDE, of the inscribed figure.

For, find (E. 1. 3.) the centre K of the circle DCLME, and join K, C and K, D and K, E: The four interior of the quadrilateral figure ACKD are (E. 32. 1. cor. 1.) equal to four right ; and (hyp. and E. 18. 8.) the ACK and ADK are right;.. the DAC + 2 CKD is equal to two right, as are also (E. 32. 1.) the three L of the isosceles ▲ CKD; .. ≤ DAC+ ≤ CKD= 2 DCK+2 CKD+4 KDC; take away the common CKD, and there remains the DAC equal to the two DCK, KDC, or to the double of the 4 KDC; because (E. 15. def. 1. and 5. 1.) the DCK 4 KDC: And, in the same manner, it

may

be shewn that the B is the double of the ▲ KDE; .. the A+B is the double of the whole CDE.

PROP. XXV.

34. THEOREM. If from any given point, in the circumference of a circle, two straight lines be drawn to the extremities of a given chord, the angle which the one makes with any perpendicular to the chord, shall be equal to the angle which the other makes with the diameter of the circle that passes through the given point.

Let C be a given point in the circumference of

the circle ABFC; let AB be a given chord; let C, A and C, B be joined; let K be the centre of the circle, and CKF a diameter passing through C; and let KD, drawn to AB, meet CB in G, Then, the KEC=L ECF.

and CA, produced, in E:

BCF, and the

For (E. 32. 1.)

EGB=

AEK+ 2 AKE=2CAK: