GEOMETRY. 
The two sides F B, B C, are' equal to the 
two sides A B, B D, and the angle DAB 
is equal F B C : the triangle A B D must 
therefore be equal to the angle F B C. But 
the parallelogram B L is double the triangle 
A B D. The square G B is also double 
the triangle F B C : consequently the paral- 
lelogram B L is equal to the square G B. 
Tlib square H C in like manner is proved 
to be equal to the parallelogram C L, which 
completes tiie solution. Euclid, 47th of 
1st Book. 
PROBLEM XVIII. 
To divide a line so that the rectangle con- 
tained under the whole line, and one segment, 
be equal to the square of the other segment. 
Fig. 21. On the given line A B describe 
the square ABCD; bisect A C in E, and 
with the distance E B extend AC to F, 
measuring from E. Make on the excess 
F A the square F H, and continue G H to 
K. The square F H will be equal to the 
parallelogram H D. 
PROBLEM XIX. 
To malte a square equal to a given right- 
lined figure. Fig. 22. Let A be the given 
right-lined figure : commute it to a parallel- 
ogram, B D, as already shown (prob. 15.) : 
add the lesser side E D to B E, so as to 
proceed to F : bisect B F in G, and from 
that point describe the semicircle B H F. 
Continue D E to H, which will give H E 
for the side of a square equal in area to the 
parallelogram B D, and to the original 
given figure A. 
PROBLEM XX. 
To find the centre of a given circle. Fig. 23. 
Draw at pleasure the chord A B, bisect it 
in D by means of a diameter, which being 
bisected will give F for the centre of the 
circle. 
PROBLEM XXI. 
To complete a circle upon a given segment. 
Fig. 24. Let A B C be the given segment : 
draw the line A C, and bisect it in D ; draw 
also the perpendicular B E through D, draw 
B A, and on it make the angle B A E, equal 
to D B A ; this will give the point of inter- 
section E for the centre, whence the circle 
may be completed. It matters not whether 
the segment be more or less than a semi- 
circle. 
PROBLEM XXII. 
To cut a given circumference into two equal 
parts. Fig. 25. Draw the line A R, bisect 
in C; the perpendicular DC will divide 
the figure into two equal and similar parts, 
PROBLEM XXIII. 
In a given circle to describe a triangle equi- 
angular to a given triangle. Fig. 26. Let 
A B C be the circle, and D E F the triangle 
given. Draw the line GH, touching the 
circle in A : make the angle H A C equal to 
D E F, and GAB equal to D F E : draw 
B C, and the triangle B A C will be similar 
to the triangle D E F. 
PROBLEM XXIV. 
About a given circje to describe a triangle 
similar to a given triangle. Fig. 27. Let 
A B C be the given circle, and D E F the 
given triangle: continue the line EF both 
ways to G and H, and having found the 
centre, K, of the circle, draw a radius, KB, 
at pleasure ; then from K make the angle 
B K A equal to D E C, and B K C equal to 
D F H ; the tangents L N perpendicular to 
K C, M N perpendicular to K B, and M L 
perpendicular to KA, will form the re- 
quired triangle. 
PROBLEM xxv. 
To describe a circle about a given triangle. 
Fig. 28. In the given triangle ABC, bisect 
any two of the angles ; the intersection of 
their dividing lines, B D and C D, will give 
the centre D, whence a circle may be de- 
scribed about the triangle, with the radius 
DC. 
PROBLEM XXVI. 
To inscribe a circle in a given triangle. 
Fig. 29. In the triangle ABC, divide the 
angles ABC, and BCA, equally by the 
lines B D, CD. Their junction at D, will 
give a point whence the circle E C F may 
be described, with the radius D F perpen- 
dicular to B C. 
PROBLEM XXVII. 
To inscribe a square in a given circle. Fi«-. 
30. Draw the diameter AC, and, perpen- 
dicular thereto, the diameter BD: the 
lines A B, B C, CD, and D A, w'ill form a 
correct square. 
PROBLEM XXVIII. 
To describe a circle at'ound a square. 
Fig. 30. In the square ABCD, draw the 
diagonals A C, B D, their intersection at E 
will give the centre of a circle, whose ra- 
dius may be any one of the four converg- 
ing lines y say EA, that will enclose the 
square. 
