AG, i. e. to the given square; and CE, BD form with ED angles equal to the given angles.

If GE does not meet the circle, the problem is impossible.

(46.) Through a given point between two straight lines containing a given angle, to draw a line such that a perpendicular upon it from the given angle may have a given ratio to a line drawn from one extremity of it, parallel to a line given in position...

Let AB, AC be the lines forming the given angle BAC, and D a point between them, and AE the line given in position. Draw any line FG parallel to AE, and take AH: FG in

E

D

M

L

B

G

the given ratio; and with the centre A, and radius AH, describe a circle, to which draw FIK a tangent. Join AI; and through D draw LMN parallel to FK, and LO parallel to FG. LN is the line required.

For AI is perpendicular to FK, and .. AM to LN; and LO is parallel to AE,

and FG : LO :: AF : AL :: (AI=) AH : AM, ;, AM : LO :: AH : FG, i. e. in the given ratio.

(47.) Through a given point between two indefinite straight lines not parallel to one another, to draw a line which shall be terminated by them, so that the rectangle contained by its segments shall be less than the rectangle contained by the segments of any other line drawn through the same point.

Let AB, AC be the given lines meeting in A. In AC take any point D, and make AE=AD. Join DE; and through I the given point draw FIG pa

rallel to DE. FIG is the line required.

M

Draw the perpendiculars FO, GO meeting in 0. Then since ED is parallel to FG, and the angles AED, ADE are equal, .. AFG and AGF are equal. But AFO=AGO, each being a right angle, .. OGF=OFG, and OF- OG; a circle ... described from the centre O, and radius OG, will pass through F, and touch AB, AC in G and F, since the angles at G and F are right angles. Let now any other line HKLM be drawn through I, and terminated by AB, AC. Since all other points in AB but G are without the circle, H is without the circle; .. HM cuts the circle in K; and for the same reason also in L. Now the rectangle KI, IL is equal to the rectangle GI, IF. But the rectangle KI, IL is less than the rectangle HI, IM; .. the rectangle GI, IF is less than the rectangle HI, IM. In the same manner it may be shewn that the rectangle GI, IF is less than the rectangle contained by the segments of any other line drawn through I, and terminated by AB, AC.

SECT. VI.

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

DA BC

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.

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

0 B D

Take a straight line AB terminated at A, and cut off 40 equal to a side of the greater, and OB equal to a side of the lesser square. With O as centre, and radius OA, describe a circle OCD; and from B draw BC at right angles to AD. The square described upon BC is the square required.

Join OC. (Eucl. i. 48.), the square described upon BC is equal to the difference of the squares on OC and OB, i. e. on AO and OB.

COR. Hence a mean proportional between the sum and difference of 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,

EB

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.

B

D

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.