GEOMETRY. 
, riven point C. Measure the angle made by 
ABC and return another of equal mea- 
surement upon the line BC, so as to make 
the angle BCD equal to ABC: the line 
C D will be parallel to the line A B. Or, 
as in fig. 11, you may from any points, say 
C D, in the line A B draw two semicircles 
of equal dimensions ; the tangent E F will 
be parallel to A B. Or you may, according 
to Problem 5, draw a perpendicular from 
the given point to the given line, and draw 
another line through the given point at right 
angles with the perpendicular proceeding 
from it to the line whose parallel was to be 
made, and which will be thus found. See 
fig. 12. 
THEOREM XT. 
Parallelograms of equal base and altitude 
are reciprocally equal. Fig. 13. 'lhe paial- 
lelogram No. 1 is rectangular: No. 2 is in- 
clined, so as to hang over a space equal to 
the length of its own base ; but the line 
A B, which is perpendicular thereto, divides 
it into two equal parts: let the left half, 
A B E, be cut off, and it will, by being 
drawn up to the right, be found to fit into 
the dotted space A C D. This theorem 
might be exemplified in various modes ; but 
we presume the above will suffice to prove 
its validity. 
THEOREM XII. 
Triangles of equal base and altitude are 
reciprocally equal. Fig. 14. As every pa- 
rallelogram is divisible into two equal and 
similar triangles, it follows that the same 
rule answers for both those figures under the 
position assumed in this proposition: we 
have shown this by fig. 15. 
PROBLEM XIII. 
To make a parallelogram equal to a given 
triangle, with a given inclination or angle. 
Fig. 16. Let B AC be the given triangle, 
and E D F the given angle. On the line D F 
measure a base equal to B C, the base of 
the triangle. Take B G equal to half the 
altitude of the triangle for the altitude of 
the parallelogram, and set it off on the line 
E D. Draw F H parallel to E D, and H E 
parallel to DF, which will complete the 
parallelogram E F D H, equal to the triangle 
BAC. 
PROBLEM XIV. 
To apply a parallelogram to a given right 
line, equal to a given triangle, in a given right 
lined figure. Fig. 17., Let A B be the given 
fine to which the parallelogram is to be an- 
nexed. Let C be the triangle to be com. 
muted, and D the given angle. Make 
BEFG equal to C, on the angle E B G : 
continue A B to E : carry on F E to K, and 
make its parallel HAL, bounded by FH, 
parallel to E A : draw the diagonal H K and 
G M both through the point B ; then K L ; 
and the parallelogram B M A L will be equal 
to the triangle C, and be situated as de- 
sired. 
problem xv. 
To make a parallelogram, on a given incli- 
nation, equal to a right-lined figure. Fig. 18. 
Let A B C D be the right-lined figure, and 
FRH the given angle or inclination ; draw 
the line D B, and take its length for the al- 
titude, F K, of the intended parallelogram, 
applying it to the intended base line K M : 
now take half the greatest diameter of the 
triangle D C B, and set it off from K to M, 
and set off half the greatest diameter of the 
triangle DAB, and set it off from H to M : 
make G H and L M parallel to F K, and 
F G parallel to K H. The parallelogram 
F K G H will be equal in area to the figure 
A B C D, and stand at the given inclination 
or angle. 
problem xvi. 
To describe a square on a given line. Fig. 19. 
Raise a perpendicular at each end of the 
line A B equal to its length ; draw the line 
C D, and the square is completed. 
THEOREM XVII. 
The square of the hypothenuse is equal to 
both the squares made on the other sides pf a 
right-angled triangle. Fig. 20. This com- 
prehends a number of the foregoing pro- 
positions, at the same time giving a very 
beautiful illustration of many. Let ABC 
be the given right-angled triangle ; on each 
side thereof make a square. For the sake 
of arithmetical proof, we have assumed 
three measurements for them : viz. the hy- 
pothenuse at 5, one other side at 4, and the 
last at 3. Now the square of 5 is @5. The 
square of 4 is 16, and the square of 3 is 9: 
it is evident the sum of the two last sides 
make up the sum of the liypothenuse’s 
square; for 9 added to 16 make 25. But 
the mathematical solution is equally simple 
and certain. The squares are lettered as 
follow: BDCE, FGBA, and AHGK. 
Draw the following lines : FC, B K, AD, 
AL, and A E. We have already shown, 
that parallelograms and triangles of equal 
base and altitude are respectively equal. 
