THEOREM 56. In any triangle the sum of the squares on two sides is equal to twice the square on half the third side together with twice the square on the median which bisecis the third side. Let ABC be a triangle, and AX the median which bisects the base BC. It is required to prove that AB2+ AC2 = 2BX2 + 2AX2. Draw AD perp. to BC; and consider the case in which AB and AC are unequal, and AD falls within the triangle. Then of the AXB, AXC, one is obtuse, and the other acute. Let the AXB be obtuse. Then from the ▲ AXB, AB2 = BX2 + AX2 + 2BX. XD. Theor. 54. And from the AXC, AC2 = XC2 + AX2 - 2XC. XD. Adding these results, and remembering Theor. 55. we have AB2 + AC2 = 2BX2 + 2AX2. Q.E.D. NOTE. The proof may easily be adapted to the case in which the perpendicular AĎ falls outside the triangle. EXERCISE. In any triangle the difference of the squares on two sides is equal to twice the rectangle contained by the base and the intercept between the middle point of the base and the foot of the perpendicular drawn from the vertical angle to the base. A ABX EXERCISES ON THEOREMS 50-53. 1. Use the Corollaries of Theorem 50 to shew that if a straight line AB is divided internally at X, then AB2=AX2+XB2+2AX. XB. 2. If a straight line AB is bisected at X and produced to Y, and it AY. YB=8AX2, shew that AY=2AB. 3. The sum of the squares on two straight lines is never less than twice the rectangle contained by the straight lines. Explain this statement by reference to the diagram of Theorem 52. Also deduce it from the formula (a−b)2=a2+b2 – 2ab. 4. In the formula (a+b) (a - b)=a2 – b2, substitute a= and enunciate verbally the resulting theorem. x+y, b x = 5. If a straight line is divided internally at Y, shew that the rectangle AY, YB continually diminishes as Y moves from X, the midpoint of AB. Deduce this (i) from the Corollary of Theorem 53 ; 6. If a straight line AB is bisected at X, and also divided (1) internally, (ii) externally into two unequal segments at Y, shew that in either AY2+YB2=2(AX2+XY2). [Euclid II. 9, 10.] case [Proof of case (i). AY2+YB2=AB2-2AY. YB =4AX2-2(AX+XY) (AX-XY) Theor. 51. Theor. 53. Case (ii) may be derived from Theorem 52 in a similar way.] 7. If AB is divided internally at Y, use the result of the last example to trace the changes in the value of AY2+YB2, as Y moves from A to B. 8. In a right-angled triangle, if a perpendicular is drawn from the right angle to the hypotenuse, the square on this perpendicular is equal to the rectangle contained by the segments of the hypotenuse. 9. ABC is an isosceles triangle, and AY is drawn to cut the base BC internally or externally at Y. Prove that AY2 AC2-BY. YC, for internal section; EXERCISES ON THEOREMS 54-56. 1. AB is a straight line 8 cm. in length, and from its middle point O as centre with radius 5 cm. a circle is drawn; if P is any point on the circumference, shew that AP2+BP2=82 sq. cm. X 2. In a triangle ABC, the base BC is bisected at X. If a = 17 cm., b=15 cm., and c=8 cm., calculate the length of the median AX, and deduce the LA. 3. The base of a triangle = 10 cm., and the sum of the squares on the other sides = 122 sq. cm. ; find the locus of the vertex. 4. Prove that the sum of the squares on the sides of a parallelogram is equal to the sum of the squares on its diagonals. The sides of a rhombus and its shorter diagonal each measure 3"; find the longer diagonal to within '01". 5. In any quadrilateral the squares on the diagonals are together equal to twice the sum of the squares on the straight lines joining the middle points of opposite sides. [See Ex. 7, p. 64.] 6. ABCD is a rectangle, and O any point within it: shew that OA2+ OC2=OB2+OD2. If AB=6.0", BC=2.5′′, and OA2+OC2=21 sq. in., find the distance of O from the intersection of the diagonals. 7. The sum of the squares on the sides of a quadrilateral is greater than the sum of the squares on its diagonals by four times the square ⚫ on the straight line which joins the middle points of the diagonals. 8. In a triangle ABC, the angles at B and C are acute; if BE, CF are drawn perpendicular to AC, AB respectively, prove that BC2 AB. BF + AC. CE. 9. Three times the sum of the squares on the sides of a triangle is equal to four times the sum of the squares on the medians. 10. ABC is a triangle, and O the point of intersection of its medians: shew that AB2+BC2+CA2=3 (OA2 + OB2 + OC2). 11. If a straight line AB is bisected at X, and also divided (internally or externally) at Y, then AY2+YB2=2(AX2+ XY2). [See p. 230 Ex. 6.] Prove this from Theorem 56, by considering a triangle CAB in the limiting position when the vertex C falls at Y in the base AB. 12. In a triangle ABC, if the base BC is divided at X so that mBX=nXC, shew that mAB2+nAC2=mBX2+nXC2 + (m +n) AX2. RECTANGLES IN CONNECTION WITH CIRCLES. THEOREM 57. [Euclid III. 35.] If two chords of a circle cut at a point within it, the rectangles contained by their segments are equal. In the ABC, let AB, CD be chords cutting at the internal point X. It is required to prove that the rect. AX, XB=the rect, CX, XD. Let O be the centre, and the radius, of the given circle. bisecting it. Join OA, OX. Proof. The rect. AX, XB = (AE + EX) (EB – EX) E Similarly it may be shewn that the rect. CX, XD = r2 – OX2. .. the rect. AX, XB=the rect. CX, XD. Q.E.D. COROLLARY. Each rectangle is equal to the square on half the chord which is bisected at the given point X. THEOREM 58. [Euclid III. 36.] If two chords of a circle, when produced, cut at a point outside it, The rectangles contained by their segments are equal. And each ectangle is equal to the square on the tangent from the point of ntersection. A In the D ABC, let AB, CD be chords cutting, when produced, at the external point X; and let XT be a tangent drawn from that point. It is required to prove that the rect. AX, XB = the rect. CX, XD = the sq. on XT. Let O be the centre, and r the radius of the given circle. Suppose OE drawn perp. to the chord AB, and therefore bisecting it. Join OA, OX, OT. Proof. The rect. AX, XB = (EX + AE) (EX – EB) = (EX+AE) (EX-AE) = EX2 - AE2 = Theor. 53. (EX2 + OE3) = (AE2 + OE2) = 0X2_OF2 - (OA2 ༩॰ ༠ And since the radius OT is perp. to the tangent XT, .. the rect. AX, XB = the rect. CX, XD the sq. on XT. = Q.E.D. |