GEOMETRY. 



merit containing an angle equal to the angle A. 

 At the point B in BC make the angle CBD equal 

 to the given angle ; describe a circle touching BD 

 in B, and passing through the point C, then the 

 segment BFC is that which is required (Prop. 

 XV. 4). 



(3.) From a given circle to cut off a segment 

 containing an angle equal 

 to a given angle. Let A be 

 the given angle ; it is re- 

 quired to cut off a segment 

 from the given circle which 

 shall contain an angle equal 

 to this angle. Take any 

 point D on the circumfer- 

 ence, and through it draw 

 a tangent to the circle ; at 

 the point D in BC make 

 the angle FDC equal to the 

 given angle, the segment 

 EGD is that which is required (Prop. XV. 4). 



THE RECTANGLE, SQUARE, AREAS, &C. 

 CONSTRUCTION (V.). 



(i.) To describe a rectangle whose two adjacent 

 sides shall be equal to two given lines. 



Take a line CD equal to A. Through D draw 

 DE perpendicular to CD, and 

 make DE equal to B. Through 

 E draw EF parallel to CD, and 

 through C draw CF parallel to 

 ED. Then CDEF is the rect- 

 angle required. The rectangle 

 CDEF is said to be contained 

 by its two adjacent sides. Hence on a given line 

 we can describe a square. 



Definition. The projection of a point upon a 

 line is the foot of the perpendicular drawn from 

 the point to the line, and the projection of one 

 line upon another is the portion of the latter 

 intercepted by the perpendiculars upon it from 

 the extremities of the former. 



PROP. XVII. If a line be divided into two 

 parts, the rectangle contained by the two parts, 

 together with the square of one of the parts, is 

 equal to the rectangle con- 

 tained by the whole line 

 and that part. 



If the line AB be di- 

 vided into two parts in *" 

 C ; it is required to prove that the rectangle 



AC CB + CB 2 = AB BC. 



From C draw CD perpendicular to AB, making 

 it equal to CB, and complete the rectangle. The 

 figure CE will be the square on CB. Then the 

 figures 



AD + CE = AE, 



or AC CB + CB 2 = AB BC 



PROP. XVIII. If a straight line be divided into 

 any two parts, the square upon the whole line is 

 equal to the squares upon each of the parts, to- 

 gether with twice the rectangle contained by the 

 parts. 



Let AB be divided into two parts in C. To 



prove that the square described on AB is equal to 



the square upon AC, and the square on CB, and 



twice the rectangle contained 



by AC and CB. On AB describe A 



the square AE, on AC the square 



AG. Produce FG and CG to 



meet BE and DE in K and 



H ; then GE is evidently the 



square on CB. Now the whole 



figure AE is made up of AG, 



GE, FH, and BG that is, the 



square on AB is equal to the square on AC and 



the square on CB, and twice the rectangle 



AC CB. 



PROP. XIX. If a straight line be divided 

 into any two parts, the square upon the whole line, 

 together with the square upon one of the parts, is 

 equal to twice the rectangle contained by the whole 

 line and that part, together with the square upon 

 the other part. 



Let AB be divided into two parts in C. To 

 prove that the square on AB, together with the 

 square on BC, is equal to the 

 square on AC, together with 

 twice the rectangle contained 

 by AB and BC. On AC de- 

 scribe the square AK ; and on 

 CB the square CF ; CK and 

 CG will evidently be in one 

 line. Produce HK to meet 

 BE in L. The figure AE, to- 

 gether with GB, is made up of 

 AK, HE, and GL. But AE is 

 the square of AB, GB is the square of CB, AK 

 the square on AC, HE is equal to GL, and each 

 is the rectangle contained by AB and BC ; there- 

 fore the square on AB, together with the square 

 on CB, is equal to twice the rectangle contained 

 by AB and BC, together with the square on 

 AC. 



PROP. XX. If a line be divided equally, and 

 unequally, the rectangle contained by the unequal 

 parts, together with the square of the parts between 

 the points of section, is equal to the square of half 

 the line. 



Let AB be divided equally in C, and unequally 

 in D. To prove that AD DB + CD 3 = CB 2 . 

 Form the rectangle AD 

 DB. Let this be AM. 

 On CD describe the 

 square CF, and on CB 

 the square CH. Pro- 

 duce EF to P, and EC A ^ 



to Q. Then AQ equals 



EH, CM equals FB, J 



therefore AM equals 

 GP and FB taken together that is, the rectangle 

 AM is equal to GP and FB together. Add to 

 these CF, which is the square of CD. Then the 

 rectangle AD DB + CD 2 = CB*. 



PROP. XXI. If a line be divided equally and 

 produced to any point, the rectangle contained by 

 the whole line thus produced and the part pro- 

 duced, together with the square of half the line, is 

 equal to the square upon the line made up of the 

 half and the part produced. 



Let AB be bisected in C, and produced to D. 

 To prove that AD DB + CB 2 = CD 2 . Form the 



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