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PROP. XLI. THEOR.

D

A

the same

ABCD

B

GEN. ENUN.-If a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram is double of the triangle.

PART. ENUN.—Let the am ABCD and the A EBC be upon

base

BC, and between the same || S BC, AE; then om

2 A EBC.
Const.–Join AC.

DEMONST.—Then the ABCD = 2 A ABC, because the diam. Ac bisects it (Prop. XXXIV.) = 2 A EBC, upon the same base BC, and between same || $. (Prop. XXXVII.)

Wherefore, if a om and a &c.-Q. E. D. Hence the areas of As are readily computed; for since the area of a om the base x its I altitude, that of a

base x the I altitude. From this Proposition the following are deductions :

m

PROP. VV. THEOR.

Gen. Enun.—The two triangles formed by drawing straight lines from any point within a parallelogram to the extremities of two opposite sides, are together equal to half the parullelogram.

PART. ENUN.--Let ABCD be a om, and E any pt. within

A.

B

E

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it. Draw EA, EB, EC, ED from the pt. E to the extremities of the opposite sides ; then A AEC + A BED,

and the A AEB + Ced, are each to half the Om ABCD.

Const. Through E draw Fer || to ac or BD. (Prop. XXXI.)

DEMONST.-1. Then the A AEC = = ŽO" ACGF, and the ABED = į Om FGDB (Prop. XLI.); .. A AEC + BED = } A ABCD (Ax. 1).

2. Subtract these =s from the whole Om ABCD; . A AEB + CED = À OM ABCD.

Wherefore the two As formed &c.-Q. E. D.

PROP. WW. THEOR.

Gen. Enun.-If two sides of a trapezium be parallel, the triangle contained by either of the other sides, and the two straight lines drawn from its extremities to the bisection of the opposite side, is equal to half the trapezium.

Part. Enun.- Let ABCD be a trapezium, having the side AB ll to the side cd. Bisect Ac in E (Prop. X.), and join EB, ED. Then Δ BED trapezium

F

B

E E

ABCD.

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Const. — Through E draw FEG || to BD, meeting cd in G, and be produced in F. (Prop. XXXI.)

DEMONST.-1. Since the alternate L FAE = alternate L ECG (Prop. XXIX.), the vertical L AEF vertical Z CEG (Prop. XV.); and, adjacent to = 28, the side AE = side EC (Const.); .. the A A EF = A ECG. (Prop. XXVI.)

2. Add to each the figure AEGDB; .. the Om FGDB = trapezium ABCD.

3. Hence the A BED = - OMFGDB, with same base and altitude (Prop. XLI.) = į trapezium ABCD.

Wherefore, if two sides &c.-Q. E. D.

CoR.-Hence a trapezium, which has two of its sides II, is = to a omformed by drawing, through the bisection of one of the sides which are not il, a line || to the opposite side, and meeting the || sides.

PROP. XX. THEOR.

A

E

Gen. Enun.—The perimeter of an isosceles triangle is greater than the perimeter of a rectangular parallelogram, which is of the same altitude with, and equal in area to, the given triangle.

Part. ENUN. and Const. Let ABC be an isosceles A.

Draw AD 1 to BC (Prop. XII.), and complete the OM ADCE, which is of the same alti. tude with the A ABC, and has all its rt. ZS. The Om ADCE =

= 2Δ. ADC (Prop. XLI.) A ABC (Props. T and NN) Then the perimeter of the A ABC is > perimeter of the Om.

DEMONST.-1. Because AB = AC, and BD = DC, .. the perimeter of A ABC 2 AC + 2 DC 2 (AC + DC); and the perimeter of the Om ADCE = 2 AD + 2 DC = (AD + DC).

2. But, because adc is a rt. _d A, Ac is > AD (Prop. XIX.); .. AC + Dc is > AD + Dc; and ... the perimeter of A ABC is > perimeter of ADCE.

Wherefore the perimeter &c.-Q. E. D.

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PROP. YY. PROB.

Gen. ENUN.—To describe a parallelogram, of which the area and perimeter shall be respectively equal to the area and perimeter of a given triangle.

Part. ENUN.-Let ABC be the gn. A; then it is required

K

D

E

G

А

B

to describe a on, whose area and perimeter shall be respectively

= to those of the A ABC.

Const.-Produce BA to D, and make AD = Ac. (Prop. III.) Bisect BD in E. (Prop. X.) Through a draw AF || to bc. (Prop. XXXI.) With cr. B, and rad. BE, describe a O cutting af in G (Post. 3). Join BG; bi. sect sc in , and through a draw he Hl to BG (Prop. XXXI.); then BGFH is the am required.

DEMONST. 1. Since HF = BG = BE (Def. 15); . . HF + BG = 2 BE = BD = BA + Ac. (Const.)

2. And since GF = BH = HC (Const.); .. GF + BH = 2 hc = BC. (Const.)

3. Hence, by addition, the perimeter of Om BGFH = perimeter of A ABC.

4. Also the Om BGFH = double of a A, upon the base BH, and between the same ll * (Prop. XLI.), A ABC (Prop. NN).

Wherefore a om has been described, whose area and perimeter the area and perimeter of the gn. A ABC.Q. E. F.

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PROP. XLII. PROB. GEN. ENUN.-To describe a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

PART. ENUN.—Let ABC be the gn. A, D the gn: rectilineal Z; then it is required to describe a om = A ABC, and having one

/ one of its XS = Z D.

Const.Bisect BC

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A

B

E

ABE

in E (Prop. X.), and join AE (Post. 1); at the pt. E in the st. line EC make Z CEF = LD (Prop. XXIII.); through a draw AFG || to bc, and through c draw cg || to EF (Prop. XXXI.); then Fecg is the on required.

DEMONST.–1. Because BE = EC, and ag is II to BC, .'. the A

A AEC (Prop. XXXVIII.); ··. the ABC= 2 AEC (Ax.9) = Om FECG. (Prop. XLI.)

2. And the am FECG has the Z CEF = the 2 D. (Const.)

Wherefore a om has been described = the gn. A ABC, and having an 2 = the gn. 2 D.-Q. E. F.

To this may be added

PROP. ZZ. PROB.

А

B

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E

Gen. ENUN.-To describe a triangle which shall be equal to a given parallelogram, and having an angle equal to a given rectilineal angle.

PART. ENUN.—Let ABCD be the gn. Om, E the gn, rectilineal Z; then it is required to describe a A Om ABCD, and having an L = LE.

Const.–At the pt. c, in the st. line dc, make the L DCF = L E (Prop. XXIII.), and let cf meet AB produced in F; produce co to G, making dG = CD (Prop. III.), and join Fg; then fcg is the A required.

Join FD.

DEMONST.-Because CD = DG, .. AFCD = A FDG (Prop. XXXVIII.); .. AFCG = 2 A FCD (Ax. 6), = Om ABCD. (Prop. XLI.)

D

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