SECT. VI.

(1.) To describe an isosceles triangle on a given finite straight line.

Let AB be the given straight line. Produce it, if necessary; and make AC and BD, each equal to one of the equal sides of the triangle. With A and B as centres, and radii AC, BD, describe circles, cutting each other in E; join AE, BE; AEB is the triangle required.

For AE=AC=BD=BE.

DA BC

(2.) To describe a square which shall be equal to the difference of two squares, whose sides are given.

Take a straight line AB terminated at

A, and cut off AO equal to a side of the eater, and OB equal to a side of the lesser

are.

With O as centre, and radius OA, describe ; and from B draw BC at right angles to are described upon BC is the square re

Eucl. i. 48.), the square described upon he difference of the squares on OC and and OB.

a mean proportional between the sum two given lines may be determined.

(3.) To describe a rectangular parallelogram which shall be equal to a given square, and have its adjacent sides together equal to a given line.

Let AB be equal to the given line. Upon it describe a semicircle ADB. From A draw AC perpendicular to AB,

and equal to a side of the given square. Through C draw CD parallel to AB, and let fall the perpendicular DE. The rectangle contained by AE, EB will be the rectangle required...

For the rectangle AE, EB is equal to the square of ED, which is equal to the square of AC, i, e. to the given square; and AB is the sum of the adjacent sides AE, EB.

(4.) To describe a rectangular parallelogram which shall be equal to a given square, and have the difference of its adjacent sides equal to a given line.

Let AB be equal to the given line. On it as diameter describe a circle. From A draw AC at right angles to AB, and .. a tangent to the circle at A; make AC equal

to a side of the given square. Take O the centre; join CO, and produce it to D. The rectangle contained by EC, CD is the rectangle required.

For the rectangle EC, CD, is equal to the square of AC, i. e. to the given square; and the difference of the sides containing the rectangle is ED=AB=the given line.

A A

(5.) To describe a triangle which shall be equal to a given equilateral and equiangular pentagon, and of the same altitude.

Let ABDCE be the given pentagon. Join AC, AD; and produce CD indefinitely both ways. Through Band E draw BG, EF respectively parallel to AD and AC. Join AF, AG. AFG is the triangle required.

=

Since AD is parallel to BG, (Eucl. i. 37.) the triangles ABD, AGD are equal; and for a similar reason, AEC AFC; .. the triangles ABD, AEC are equal to AGD, AFC; to these equals add the triangle ADC; and the pentagon ABDCE is equal to the triangle AGF; and they have the same altitude, viz. the perpendicular from A upon DC.

(6.) To describe an equilateral triangle equal to a given isosceles triangle.

Let ABC be the given isosceles triangle. On AC describe an equilateral triangle ADC, and from D draw DE perpendicular to AC; it will also bisect AC and pass through B. On DE describe a semicircle; and from B draw

BF perpendicular to DE, meeting the circle in F. with the centre E, and radius EF, describe a circle meeting ED in G; draw GH, GI parallel to DA, DC respectively; the triangle GHI is equilateral, and equal to ABC.

Since GH is parallel to AD, and GI to DC, the triangles GHI, ADC are similar; but ADC is equilateral, and.. also GHI is equilateral.

Also (Eucl. vi. 8. Cor.) ED: EG

EG: EB, and (Eucl. vi. 2.) ED : EG :: EA : EH, .. EG EB :: EA: EH,

and .. (Eucl. vi. 15.) the triangles EGH, EBA are equal. But GHE=GIE, and BAE= BCE, .. also GHI=B4C.

(7.) To describe a parallelogram, the area and perimeter of which shall be respectively equal to the areá and perimeter of a given triangle.

B

E

F

Let ABC be the given triangle. Produce AB to D, making BD=BC; bisect AD in E; draw BF parallel to AC; and with the centre A, and radius AE, describe a circle cutting BF in G. Join AG; and bisect AC in H. Draw HF parallel to AG. AGFH is the parallelogram required.

H

For HF-AG=AE, .. HF and AG together are equal to AD, i. e. to AB and BC together; and GF= AHHC,.. the perimeter of AGFH is equal to the perimeter of ABC; and AGFH is double of a triangle on the base AH and between the same parallels, and .. is equal to the triangle ABC.

(8.) To describe a parallelogram which shall be of given altitude, and equiangular and equal to a given parallelogram.

1

Let ABCD be the given parallelogram, and EF the given altitude. Draw EH and FG at right angles to FE; and at the

D

B

H

point F, in the line GF, make the angle GFI equal to CDA; take FG a fourth proportional to FI, AD and DC; and from G draw GH parallel to FI, meeting EH produced in H; IFGH is the parallelogram required.

=

For its altitude is EF; and the angle GFI CDA, :. FIH=DAB; whence the parallelograms are equiangular; and they are equal; since the sides about the equal angles are reciprocally proportional (Eucl. vi. 14.).

(9.) To describe a square which shall be equal to the sum of any number of given squares."

Let AB be a side of one of the given squares. From B draw BC perpendicular to AB, and equal to a side of the second square. Join AC; and from C draw CD perpendicular to it, and equal to a side of A the third square. Join AD; and from D draw DE perpendicular to AD, and equal to a side of the fourth. Join AE. The square of AE is equal to the squares of AB, BC, CD, DE.

Since the angles ADE, ACD, ABC are right angles, the square of AE is equal to the squares of AD, DE, i. e. to the squares of AC, CD, DE; and .. to the squares of AB, BC, CD, DE.

And by proceeding in the same manner whatever be