JuLV 11, 1884.] 



♦ KNO\A/'LEDGE • 



41 



siiuaro on A D equal to the square on AC, and ri'^jlit times the 

 square on D C ; that ig, to nine times tlie square on A C. Hence if 

 a strafilit line be divided into three equal parts, the square on the 

 syhole lino is equal to nine times the square on any one of the 

 parts. 



Prop. 9 thus, — Let the straight line A 15 be divided into two equal 

 parts at the point C, and into two unequal parts at the point D ; 

 then tlie squares on r 



A U, D Bare together A S ^ B 



equal to double the 

 squares on AC, CD. 



The square on AD is equal to the squares on .\ C, C D, together 

 with twice the rectangle A C, C D ; that is, the square on A D is 

 iireater than the squares on C B, C D by twice the rectangle 

 C B, C D. And the square on D B is less than the squares on 

 C B, C D by twice the rectangle C B, C D (Prop. 7, 2nd enunciation). 

 Hence, adding the squares on A D and D B are together equal to 

 double the squares on C B, C D. 



Prop. 10 thus, — g ^ C B o 



Let the straight line 



.K B be bisected in C, 



and produced to D ; then the squares on A D, D B shall be together 



double of the squares on .\ C, (' D. 



Produce D A to E making A E equal to B D ; so that E C is equal 

 to C D and E B to A D. Hence, by the preceding proposition, the 

 squares on E B, B D are together double the squares on E C, C B. 

 That is, the squares on A D, B D are together double the squares on 

 C D, C B. 



Props. 9 and 10 may be included under one enunciation thus, — 



The sc[iiares on tu-o liner (A I) anii D B) are tor/ether double the 

 squares on half the sum and half the difference of the tiuo lines ; 



or thus 



The squares on the sum and difference of two lines (AC and C D) 

 are together double the sq^iarei on the tv:o lines. 



Cor. — Since the squares on AD, D B exceed the squares 

 on A C, C B by twice the square on C D, it follows that when a 

 straight line is bisected the sum of the squares on the two parts is 

 least, and the sum is greater as the difference between the two 

 parts of the divided line is greater. 



(To be continued.) 



2AST KIDERS OS EUCLID'S FIRST BOOK. 



With Suggestions. 



(Continued from page 19.) 



149. If the opposite sides of a quadrilateral figure are equal the 

 figure is a parallelogram. 



150. If the opposite angles of a quadrilateral figure are equal the 

 figure is a parallelogram. 



151. The two straight lines AB, AC, intersect in A, and P is a 

 point within the angle BAG. It is required to draw a straight line 

 B P C so that B P may be equal to P C. 



Suppo.-^e B P equal to PC ; j"in A P and produce to D so that P D 

 mai/ be equal to A B. Then A B D C t.< a parallelogram, ij"c. 



152. With the same construction, Q is a point without the angle 

 BAG. It is required to draw Q B C so that Q B may be equal to 

 BC. 



Take a point E in A B produced, so that B E may be equal to A B. 

 Then Q E C A i.-^ a parallelogram (if Q B be assumed equal (o B C. 



153. From a given point in one of two intersecting lines it is 

 required to draw a line terminated by the second, and such that 

 the line drawn from the jioint of intersection of the given lines to 

 the bisection of the required line may make given angle with one 

 of the given lines. 



154. From a given point P it is required to draw three straight 

 lines P A, P B, and P C equal respectively to three given straight 

 lines, and having their extremities A, B, and C in one straight line 

 and A B equal to B G. 



Suppose the lines drawn as required P B lying between P A and 

 P C ; then if P B be produced to D so that B D is equal to P B, 

 P A D C is a parallelogram, A-c. 



155. Draw a straight line through a given point such that the 

 part of it intercepted between two given parallels may be of a 

 given length. 



156. Draw a straight line through a given point lying between 

 two parallels, so that the line may be terminated by the parallels, 

 and divided by the given point into two parts having a given dif- 

 ference. 



157. If the diameters of a quadrilateral figure bisect the angles 

 the figure is a rhombus. 



158. If one diameter of a parallelogram bisect opposite angles 

 it is a rhombus. 



159. If the diameter of a parallelogram intersect at right angles 

 it is a rhombus. 



160. Straight lines bisecting adjacent angles of a parallelogram 

 intersect at right angles. 



161. Straight lines bisecting opposite angles of a parallelogram 

 having unequal sides are parallel to each other. 



162. If the diameters of a parallelogram are equal it is a rect- 

 angle. 



163. If the diameters of a quadrilateral figure bisect the angles 

 and are equal the figure is a square. 



164. Find a point such that the perpendiculars let fall from it 

 on two given straight lines may be equal to one another. 



163. Between two given straight lines draw a straight line which 

 shall be equal to one straight line and parallel to another. 



166. On A B a side of the parallelogram A B G D a parallelogram 

 A F E B is described, so that E B is in the same str.aisiht line with 

 B D, and F B with B G. Show that E B is equal to B D. 



167. Equilateral triangles are described on the four sides of a 

 parallelogram. 



168. On A B, DC opposite sides of a parallelogram, equilateral 

 triangles ABE and C F D are described towards the same parts ; 

 show that F E A D and F E B C are parallelograms. 



169. If in Ex. 168, C F D and A E B are described towards 

 opposite parts, then D E B F and C E A F are pai-allelograms. 



170. From A, C, opposite angles of the parallelogram A B C D, 

 are drawn the four lines, A F, A E, G G, G H, perpendicular respec- 

 tively to the sides A D, A B, C B, and C D, and on the side remote 

 from the parallelogram ; also A F is equal to G G, and A E to G H. 

 Show that E G H F is a parallelogram. 



171. Equilateral triangles are described on the four sides of a 

 parallelogram. Show that the vertices of these triangle fall on the 

 angles of a parallelogram, — 



(i.) When all the triangles are towards the same parts as the 



parallelograms. 

 (ii.) When all the triangles are towards opposite parts, 

 (iii.) When two triangles on opposite sides are towards the 



same parts, and the other two triangles towards opposite 



parts. 



172. On the sides A B, B C, and G D of a parallelogram A B C D 

 three equilateral triangles ABE, B C F, and C D G are described, 

 ABE and C D G towards the same parts as the parallelogram and 

 B F C towards opposite parts. Show that E F and F G are respec- 

 tively equal to two diagonals of the parallelogram. 



Shov: that the triangle BFE is equal in all respects to the 

 triangle ABC. 



173. Show that the same holds good if BFC lies towards the 

 same parts as the parallelogram and A B E, G D G, towards opposite 

 parts. 



174. In the parallelogram A B C D, the angle A D B is equal to 

 the angle A C B. Show that A B G D is rectangular. 



(To be continued.) 



(But Cfjtss Column. 



Br Mephisto. 



CHESS GOSSIP. 



A TESTIMONIAL is now being raised for England's Chess 

 Champion, Mr. Blackbume, who has recently undergone a 

 severe illness. 



The Field devotes a leading article in discussing the desirability 

 of having a National Chess Association. That such an institution 

 would benefit the cause of Chess goes without saying. 



The same writer condemns the proceedings of the Counties Chess 

 Association. It is quite unworthy of amateur Chess players to 

 appeal to the public for funds, and at the same time exclude those 

 players in whose performances the public take the most interest. 



We are afraid the secretary betrayed the weakness of his cause 

 when he thought desirable to put forth the names of MacDonnell 

 and Bird as likely competitors, with a general promise to admit 

 first-rates. It may also be reckoned a very clever move to exclude 

 first-rates, after once having attracted the notice of the public by 

 the bait. 



The following story, which comes from an eye-witness, may 

 throw some light upon the matter. 



Scene : a meeting of the Counties Chess Association. A promi- 

 nent member of CCA. playing with a fairly strong player from 

 London. Prominent member, being in diSiculties, takes half an 

 hour to consider his move. Time limit twenty moves an hour. 

 London player politely (but with just a suspicion of sarcasm) : 

 '' Your sand is running out, sir." Pronunent player of the CCA. 



