BOOK III. 22. EXERCISES. [Let ABCDEF... be the polygon and O the centre of the circle. The ABC 2 rt. 4° - AOC; ¿CDE=2 rt. 4° - ¿COE, etc.; .. on addition, the sum of the alternate angles - =2 rt. 4° x no. of the sides - all the 4 at O=etc.] 149 6. Shew that the four straight lines bisecting the interior (or the exterior) angles of any quadrilateral form a quadrilateral which can be inscribed in a circle. [Let the bisectors of the interior angles A and B, B and C, C and D, D and A of a quadrilateral meet in E, F, G, H. FEH=2 rt. 2° - 1 LA-4B, and FGH 2 rt. 4° -14C-4D; Then <*FEH, FGH=4rt. 2o - half the sum of the interior angles of ABCD =4rt. 4-2 rt. 2o2 rt. 48; .. etc. Similarly for the bisectors of the exterior angles.] **7. Shew that no parallelogram except a rectangle can be inscribed in a circle. 8. D is any point on the arc BC of a circle whose centre is A; CD is produced to E; prove that the angle BDE is half the angle BAC. 9. AOB is a triangle; C and D are points in BO and AO respectively, such that the angle ODC is equal to the angle OBA ; shew that a circle may be described round the quadrilateral ABCD. 10. ABCD is a quadrilateral inscribed in a circle, and the sides AB, DC when produced meet at O; shew that the triangle AOC is equiangular to the triangle BOD, and the triangle AOD to the triangle BOC. **11. If any two consecutive sides of a hexagon inscribed in a circle be respectively parallel to their opposite sides, the remaining sides are parallel to each other. [Let ABCDEF be the hexagon having FA, AB parallel to CD, DE. Then FEB 2 rt. 23 - 2 FAB [III. 22]=2 rt. 2o - DEC [I. 29]=▲ EBC. 12. ABCD is a quadrilateral inscribed in a circle; AB and CD meet in P; AD and BC meet in Q; prove that the bisectors of the angles P and Q are at right angles. [If O be the intersection of the bisectors, and OQ meet CD in L, then 4 OPC D-PAD and OQD={D}DCQ; :. ¿POQ=¿PLQ-2OPC=4D-4 OQD - 4 OPC =2D-(24D-PAD-PCQ) BOOK VI. 31. 273 PROPOSITION 31. THEOREM. In any right-angled triangle, any rectilineal figure described on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. Let ABC be a triangle, right-angled at A: the rectilineal figure described on BC shall be equal to the similar and similarly described figures on BA and CA. Let K, L, M be the similar and similarly described figures on BC, CA, and AB. Proof. Because AD is drawn from the right angle A perpendicular to BC, the ACBA is similar to the ▲ ABD; ... BC: BA: BA: BD; [VI. 8. [VI. Def. 2. [VI. 20, Cor. 2. [V. B. Similarly, CD: BC :: .. as BD and CD together are to BC so are L and M together to K. But BD and CD together = BC; .. the figure K = the figures L and M. Wherefore, in any right-angled triangle, etc. [V. 24. [V. A. [Q.E.D. Note. Of this general proposition I. 47 is a particular case. xxxii EUCLID'S ELEMENTS. EXERCISES. **1. The straight line joining two points, A and B, is the polar of the point of intersection of the polars of A and B. [Let the latter intersect at T; then A lies on the polar of T, since T lies on the polar of A (43); so B lies on the polar of T; .. AB is the polar of T.] **2. The point of intersection of any two straight lines is the pole of the straight line joining their poles. **3. Find the locus of the poles of all straight lines which pass through a given point. [Use Art. 43.] **4. A and B are two points in a plane of a circle whose centre is C; AX and BY are the perpendiculars from A and B on the polars of B and A respectively; prove that the rectangles CA. BY and CB. AX are equal. [Salmon's Theorem.] [Let the polar of A meet CA in M, and that of B meet CB in N; also draw AU, BV perp' to CB, CA respectively. Then A, V, U, B lie on a circle; . CU. CB=CA. CV. [III. 36. But CN, CB=CM. CA=sq. on radius (Art. 42). Subtract; .. CB.NU=CA . MV, i.e. CB. AX-CA. BY.] 44. Orthogonal Circles. Def. Two circles are said to intersect orthogonally when the tangents at their points of intersection are at right angles. If the two circles intersect at P, the radii O̟P and ОP, which are perpendicular to the tangents at P, must also be at right angles. i.e. the square of the distance between the centres must be equal to the sum of the squares of the radii. Also the tangent from O, to the other circle is equal to the radius ɑ2, i.e. if two circles be orthogonal the length of the tangent drawn from the centre of one circle to the second circle is equal to the radius of the first. Either of these two conditions will determine whether the circles are orthogonal. It follows that if we want the circle whose centre is O, which shall cut a given circle, centre O1, orthogonally, we must take its radius equal to the tangent from O2 to the given circle. lxxxvi EUCLID'S ELEMENTS. COMPLETE QUADRILATERALS. 106. Let ABCD be an ordinary Euclidean quadrilateral. Let AD The straight line PQ is called the third diagonal, the other two Also, if AC, BD meet in R, then P, Q, and R are called the three 107. The middle points of the diagonals of a complete quadrilateral lie B Complete the parallelograms QASC, QBTD. Then PC: CU :: PD: DA, since CD, UA are |}, .. PC: PV:: CU: VB, :: CS VT, since CUS, VBT are similar A; .. PST is a straight line; the middle points of QP, QS, QT are in a straight line parallel to [Art. 24. But since QBTD, QASC are gms, the middle points of QT, QS are .. the middle points of the three diagonals BD, AC, QP lie on a Please state which Part is required Book I., 1s.; Books I. and II., 1s. 6d.; Books III. and IV., EUCLID'S 2s.; ELEMENTS OF GEOMETRY BOOK I. BOOKS I. AND II. BOOKS III. AND IV. EDITED FOR THE USE OF SCHOOLS BY CHARLES SMITH, M.A. MASTER OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE AND SOPHIE BRYANT, D.Sc. HEAD MISTRESS OF THE NORTH LONDON COLLEGIATE SCHOOL FOR GIRLS London MACMILLAN AND CO., LIMITED NEW YORK: THE MACMILLAN COMPANY All rights reserved |