i. e. in the duplicate ratio of BE BE BF, i. e. in the ratio of n

BG, or the ratio of

n-1,

.. the gnomon AKC is to AC in the ratio of 1: n.

(54.) To describe a triangle equal to a equal to a given rectilinear figure, having its vertex in a given point in a side of the figure, and its base in the base (produced if necessary) of the figure.

Let ABCDEF be the given rectilineal figure, and P a given point in CD, which is to be the vertex of the triangle, the base being in AF. Join CA, and draw BG

parallel to it; join CG, PG, PF, PE. Draw CH parallel to PG. Join PH. Draw DI parallel to PE, meeting FE produced in I. Join PI; and draw IK parallel to PF, meeting AF in K. Join PK; HPK will be equal to ABCDEF.

Since BG is parallel to CA, the triangles BAG, BCG are equal; the figure therefore is equal to GCDEF. And since GP is parallel to CH, the triangles GCP, GHP are equal. Again since DI is parallel to PE, the triangles PIE, PDE are equal; .. PDEF is equal to the triangle PIF, i. e. to the triangle PKF, since IK is parallel to PF; whence the whole figure

E E

ABCDEF is equal to the triangles PHG, PGF, PKF, i. e. to the triangle PHK.

(55.) On the base of a given triangle, to describe a quadrilateral figure equal to the triangle, and having two of its sides parallel, one of them being the base of the triangle; and one of its angles being an angle at the base, and the other equal to a given angle.

Let ABC be the given triangle, AC its base. At the point C make the angle ACD equal to the given angle; b and let CD meet BD drawn parallel to AC, in the point D. On BD describe a semicircle BED; draw AF parallel

H

GF

to CD, and FE perpendicular to BD; and with the centre D, and radius DE, describe the arc EG. Draw GH parallel to AF, and HI to AC; AHIC will be the figure required.

Join HC. Since DG = DE,

BD: DG :: DG: DF,

..(Eucl. v. 19.) BG: GF :: DG: DF:: HI: AC. Now BG GF: BH: HA,

.. BH : HA :: HI : AC.

But the triangles HCI, AHC are in the proportion of HI: AC, and the triangles BHC, AHC in the proportion of BH: HA,

=

:. HCI : AHC :: BHC : AHC,

or HCI BHC; .. ACH, and HCI together are equal to ACH, and BCH together, or AHIC=ABC.

(56.) A trapezium being given, two of whose sides are parallel; to describe on one of those sides another trapezium, having its opposite side also parallel to this, and one of the angles at the base the same as the former, and the other equal to a given angle.

Let ABCD be the given trapezium whose sides AD, BC are parallel. Join BD; and draw CE parallel to

produced in E.

it, meeting AB

Then the trian

E

F

B

gles BCD, BED are equal; and .. the triangle AED is equal to ABCD. Hence (vi. 55.) a figure ADGF a figure be described equal to ADE, and.. to ADCB.

may

(57.) If with any point in the circumference of a circle as centre, and distance from its centre as radius, a circular arc be described; and any two chords be drawn, one from the centre of the circular arc, and the other through the point where this cuts the arc, and parallel to the line joining the centres; the segments of each chord intercepted between the circumferences which are concave to each other, will be equal respectively to those of the other between the other circumferences.

With any point C in the circumference of the circle ABC as centre, and radius CE equal to the distance from the centre E, let a circle DFE be described. Join CE, and

FH, and GF=FA.

H B

draw any chord CFA; and through F draw HFG parallel to CE; then will CF Produce CE to B,

=

and join HE. And since HG is

parallel to BC, the angle FHE is equal to HEB. Also since the circles are equal, the arc HB is equal to the arc FE (ii. 1. Cor. 1.), .. the angle HEB is equal to FCE, :. FCE = FHE, and HC is a parallelogram; whence HF=EC=CF. Also since the rectangle CF, FA is equal to the rectangle HF, FG, and HF=CF, .. FA=FG.

COR. Hence if any number of lines be drawn parallel to BE, and terminated by the two circumferences, each of them will be equal to BE.

(58.) If two diagonals of an equilateral and equiangular pentagon be drawn to cut one another, the greater segments will be equal to the side of the pentagon; and the diagonals cut one another in extreme and mean ratio.

B

Let ABDCE be an equilateral and equiangular pentagon; draw the diagonals E ED, BC cutting each other in F; EF and FB will be each equal to a side of the pentagon; and ED, BC are cut at F, in extreme and mean ratio.

About the pentagon describe a circle. And since AB=CE, the arcs AB, CE are equal; .. AE is parallel to BC. For the same reason, AB is parallel to EF;.. the figure ABFE is a parallelogram; whence AB=FE, and AE=FB; but AB=AE, «. EF=FB, and each is equal to a side of the pentagon.

Also the angle DCF=CDF=DEC,.. the triangles DCF, DEC are similar,

and ED CD :: CD: DF,

or ED: EF :: EF : FD ;

.. ED is cut in extreme and mean ratio. The same may also be proved of BC.

(59.) If the sides of a triangle inscribed in the segment of a circle be produced to meet lines drawn from the extremities of the base, forming with it angles equal to the angle in the segment; the rectangle contained by these lines will be equal to the square described on the base.

D

Let the sides AB, CB of the triangle ABC, inscribed in the segment ABC, be produced to meet CE, AD, which make with AC, angles equal to the angle ABC in the segment; the rectangle AD, CE is equal to the square of AC.

A

Since the angle ABC=DAC, and the angle at C is common to the triangles ABC, ADC, the triangles are similar. In the same manner it may be shewn that ABC, AEC are similar; and .. ADC, AEC are also similar; whence

AD: AC AC: CE,

and the rectangle AD, CE is equal to the square of AC.

(60.) If two triangles (one of them right angled) have the same base and altitude, and the hypothenuse intersect a line which is drawn bisecting the right angle; a