220° 
Problems on the hypothenuse isa proportional between DE and» 
to Sect. IV. E)F, (21.) that is between A and B. 
—  Lisy 
Fig. 122. 
Fig, 123. 
Fig. 124. 
Fig. 125. 
Fig. 126, 
Hes. orig 
- Pros. IV. 
To divide a straight line AB into two parts, so that 
one of them shall be a mean proportional between the 
whole line and the other part. , 
AtB, one extremity of the line, erect a perpendicular 
BC equal to half AB; on C asacentre, with CB asa 
radius, describe a circle; draw AC to meet the cir- 
cumference in D, and take AF = AD, then the line 
AB shall be divided at the point F in the manner re- 
quired, that is, AB: AF :: AF: FB. ! 
For AB being perpendicular to the extremity of the 
radius BC, is a tangent to the circle, (9. 2.) ; there- 
fore if AC be produced until it meet the circumference 
againin E, we shall have AE: AB :: AB; AD; hence, by 
division, AE—AB:AB::AB—AD:AD; but since BC 
=1.AB, therefore DE = AB, and consequently AE — 
AB=AD = AF; also, because AF = AD, we have 
AB—AD = FB; therefore AF: AB:: FB: AD, or 
AF, and by inversion, AB: AF: : AF: FB, 
Scnotium. A line divided in this manner is said to 
be cut in: extreme and mean ratio; and it may be re- 
marked, that AE is alsodivided into extreme and mean 
ratio at the point D, for since AB= DE, we have 
AE; DE:: DE: AD. 
Pros. V. 
Through a given point A in a given angle BCD, to 
draw a straight line BD; so thatthe parts AB, AD 
contained between the point A and the two sides of 
the angle shall be equal. 
Through the point A, draw AE parallel to CD ; 
take EB = EC, and draw AB to meet CD in D, and 
the thing is done. For AE being parallel to CD, we 
have BE: EC::BA: AD; but BE= EC, therefore 
BA = AD. 
Pros, VI. 
To make a square equivalent to a given parallelo- 
gram, or to a given triangle. 
1. Let ABCD be the given parallelogram, AB its 
base, and DE its altitude. Between AB and DE find 
a mean proportional XY (Prob. 3.); the square made 
on XY shall be equivalent to the parallelogram ABCD. 
For since by construction, AB : XY::XY: DE; there- 
fore, (Cor. 22.) XY°=AB x DE = the parallelogram 
ABCD, (Cor. '1.) 
2. Let ABC be the given triangle, BC its base, 
and AD its altitude. Take a mean proportional be- 
tween BC and the half of AD; and let XY be that mean. 
The square made on XY shall be equivalent to the tri- 
angle ABC, 
or since BC: XY:: XY: 4AD; therefore XY? = 
BC x 4} AD = triangle ABC. 
Pros, VII. 
Upon a given straight line AD, to make a rectangle 
DAEX equivalent to a given rectangle ABFC. 
Find AX a fourth proportional to the three lines AD, 
GEOMETRY: . Race Hg 
| AX=AB x AG (22.); therefore the 
AB, AC, (Prob. 2.); the rectangle contained by AD. Prot 
and AX willbe thatrequired. | i ge tee 
cE ee ae ene — 
rettangle ADE} 
is equivalent to the rectangle ABFC. 
Pros. VIII.) | * anne 
Having given any rectilineal figure to make another 
equivalent to it, that shall have one side fewer. — ~ 
Let ABCDE be the given figure. Draw a diagonal Fig. 
CE so as to cut off from it a tri CDE; erie 
D, the vertex of the triangle, draw DF parallel. ry 
a ee ee 
figure (produced if necessary) in F ; join CF,. : 
figure ABCF shall be equal to the figure ABCDE,, 
and have one side fewer. . For the ce CEE is: 
equivalent to CDE (6.), therefore, addi common 
space ABCE, the figure ABCF is equal to the figure 
ABCDE. POPPN BS, © le “a 
Scuoium,. By this problem, a trianglema pred 2 
that shall be cealehe to a rectilineal nig ;.= 
number of sides. Thus, the five-sided Sure ABCDE 
having been reduced to the quadrilateral ABCF, if we 
join CA, and draw BG parallel to CA, to meet FA 
produced in G, and then join CG, the 
will be transformed into an equivalent triangle CGF, 
which will also be equal to the original figure ABCDE. 
It_ has been already shewn that, a square may b 
found equivalent to any triangle; therefore by this, an 
Prob. 6. any rectilineal figure whatever may be trans- 
formed into an equivalent square. Sa 
Pros. IX. 
To make a square that shall be equal to the sum, or 
to the difference of two given squares. 
* 
PP 
Let A and B be the sides of the two given squares. pig. 195 
1. To make a square equal to the sum of two squares, { 
draw two indefinite lines ED, EF, containing a Hight ~ 
angle; take DE=A and EG=B; join DG, and be 2 
is evidently the side of the square sonpned. i) 
2. Ifa square is to be found equal to the difference 
of two squares; form a. right angle FEH; take EG 
equal to the lesser of the two sides A, B, and on G as La 
a centre, with a radius equal to the greater, describean = _ 
arc, to meet EH in H; then EH shall be the side of a By 
square equal to the difference of the squares on GH 
and GE, or on A and B, as is evident from Prop. 11. 
Sect. 4. ‘ 
Scuotium. By this problem, a square may be made 
equal to any number of given squares. 
Pros. X. 
On a given straight. line FG, homologous to AB, to Fig 
deaieibh's polygon similar to a given polygon ABCDE. 
Draw the diagonals AC, AD: at the point F make — 
the angle GFH equal to the angle BAC, and af G 
make the angle FGH equal to ABC; the lines FH, 
GH will meet in H, and form a triangle FGH similar to 
ABC. Inlike manner, on FH, which ishomologous 
to AC, construct a triangle FHI, similarto ACD; and 
on FI, homologous to AD, construct)a triangle FIK 
similar to ADE. The polygon FGHIK shall be simi- 
lar to ABCDE. For hese two polygons are composed 
