CHAMBERS'S INFORMATION FOR THE PEOPLE. 



rectangle AD DB. Let 

 this be AM. On CB and 

 CD describe the squares 

 CF and CH. Produce 

 EF and EC to meet HM 

 and NM respectively in P 

 and Q. Then AQ equals 

 BP, CM equals GP ; there- 

 fore AM equals GP and 



BP taken together. Add to these CF, which is 

 the square on CB, then AD DB + CB a = CD'. 

 Hence it follows that the difference of two squares 

 is a rectangle, one of whose sides is the sum of 

 the sides of the two squares, and the other is their 

 difference. 



PROP. XXII. In a right-angled triangle, the 

 square on the side opposite the right angle is equal 

 to the sum of the squares described on the sides 

 containing it. 



Let ABC be a right-angled triangle, having the 

 angle at A a right angle ; it is required to prove 

 that the square on BC is 

 equal to the square on BA, 

 together with the square 

 on AC. On the sides BC, 

 BA, AC describe the 

 squares BE, BG, and AK. 

 Join AD and FC, and 

 through A draw AL paral- 

 lel to BD. Then it is 

 evident that the two tri- 

 angles ABD, FBC are 

 equal, as they have two 

 sides in the one equal to 

 two in the other, and also 



the contained angles equal ; their doubles must 

 therefore be equal that is (Prop. XI. Cor. 3), BL 

 must be equal to GB. In a similar manner it 

 may be shewn that CL is equal to AK ; there- 

 fore the whole square BE is equal to the two 

 squares BG and CH taken together. 



PROP. XXIII. If through a point in the plane 

 of a circle a line be drawn to cut the circumference, 

 the rectangle contained by the two distances of the 

 faced point front the points of intersection is always 

 the same, however the line may be drawn. 



(i.) If the point P be within the circumference. 

 Through P draw any line SPT, 

 cutting the circumference in 

 S and T ; it is required to 

 prove that the rectangle SP 

 PT is always the same, how- 

 ever the line may be drawn. 

 From the centre O draw the 

 line OQ perpendicular to the 

 chord ST, which it will bisect 

 (Prop. XII.). Then since the 

 chord is bisected in Q, and divided unequally in 

 P, we have (Prop. XX.) 



SP PT + PQ 2 = QT 2 . 

 Add to each OQ 2 ; 



.-. SP PT + PQ 2 + OQ 2 = QT 2 + OQ 2 , 

 or SP PT + (Prop. XXII.) OP 2 



= (Prop. XXI I.) (radius) 2 ; 

 .'. SP PT = (radius) 2 OP 2 . 



Now OP is the distance of the point from the 

 centre ; and since the point is fixed, this cannot 



620 



vary, and the radius of the circle does not vary ; 

 hence the rectangle SP PT is always the 

 however the line through P may be drawn. 



(2.) If the point P be without the circumfere 

 Through P draw any line 

 cutting the circle in S and T ; 

 it is required to prove that the 

 rectangle SP PT is always 

 the same, however the line 

 may be drawn. From the 

 centre O draw the line OQ 

 perpendicular to ST, which it 

 will bisect (Prop. XII.). Then 

 since ST is bisected in Q, and 

 produced to P, we have (Prop. 

 XXI.) SP PT + TQ 2 = PQ 2 . 



Add to each OQ 2 ; 



.-. SP PT + TQ 2 + OQ 2 = PQ 2 + OQ 2 , 

 or SP PT + (radius) 2 = OP 2 (Prop. XXII.) ; 

 .'. SP PT = OP 2 - (radius) 2 . 



But, as before, since the point is fixed, its dis- 

 tance from 'the centre cannot vary ; hence the 

 rectangle SP PT is always the same, however 

 the line through P may be drawn. 



Cor. If from P the tangent PM be drawn, it 

 follows that 



SP PT = PM PM, 

 or SP PT = PM 2 ; 

 or that, if from a point with- 

 out the circumference a line 

 be drawn cutting the circum- 

 ference, and another touching 

 it, the rectangle by the whole 

 line cutting the circumference 

 and the part without it is 

 equal to the square of the 

 line touching the circumference. 



Conversely, it may be shewn that if SP PT 

 = PM 2 , PM will be a tangent to the circle. 



CONSTRUCTIONS (VI.). 



(i.) To describe a square equal to the sum of 

 two squares. 



Let AB and C be the sides 

 of the two given squares. 

 From the extremity of AB 

 draw a line BD perpendicular 

 to it and equal to C. Join 

 AD. Then the square on AD is equal to the sum 

 of the squares on AB and C (Prop. XXII.). 



(2.) To describe a square equal to the difference 

 of two squares. 



Let AB and C be the sides 

 of the given squares. On AB 

 describe a semicircle. In this 

 semicircle inscribe a chord 

 AD equal to C. Join BD. 

 Since (Prop. XV. 2) ADB ' 

 is a right angle, the square on BD is evidently 

 equal to the difference of the squares on AB and C. 



(3.) To describe a square equal to a given 

 rectangle. 



Let AC be the given rect- 

 angle ; it is required to de- 

 scribe a square which shall be 

 equal to it. Produce AB to 

 D, making BD equal to BC, 

 and on AD as diameter de- 

 scribe a semicircle. Produce 

 CB to meet the circumference in E. Then the 



