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(43.) Given the base, one of the angles at the base, and the difference between the side opposite to it and the perpendicular ; to construct the triangle.

Take an indefinite line AB; and from any point C in it draw CD perpendicular to it, and equal to the given difference. From D draw DA making the angle DAC equal to the given angle. Draw DE parallel to AB, and equal to the given base. Join AE; to which from D draw DF=DC. Draw EG parallel to FD, and meeting AD produced; EDG will be the triangle required.

From G draw GB perpendicular to AB. Since DF is parallel to GE, and DC to GB,

DC: GB: AD: AG :: DF: GE. But DC being equal to DF, GB is equal to GE; whence the difference between GE and GH is equal to HB, or DC, i.e. to the given difference. And the angle GDE is equal to DAC, i. e. to the given angle; and DE is equal to the given base.

(44.) Given the vertical angle, the difference of the base and one side, and the sum of the perpendicular drawn from the angle at the base contiguous to that side upon the opposite side and the segment cut off by it from that opposite side contiguous to the other angle at the base ; to construct the triangle.

Let AB be equal to the given sum, and BC to the given difference; and let them be placed so as to contain

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an angle ABC, which with the given vertical angle will be equal to two right angles. Through C draw DCE parallel to AB, and through B draw DBF making half a right angle with it. Join AD; and to it from B draw BG equal to BC; and parallel to these respectively draw AF, FH; AFH will be the triangle required.

Produce FH to l; and let fall the perpendicular FK. Since BC is parallel to FI, and BG to AF,

BC: FI :: BD : FD :: BG : AF, but BC=BG, .. FI=AF; and HI=BC will be the difference between the base AF and the side FH. Also since the angle BFK is half a right angle, FK=KB; and AB is the sum of the perpendicular FK and the segment KA. Also BCIH being a parallelogram, the angle AHF, or its vertically opposite IHB, together with ABC will be equal to two right angles ; and .. AHF is equal to the given vertical angle.

(45.) Given the base, the difference of the sides, and the segment intercepted between the vertex and a perpendicular from one of the angles at the base upon the opposite side ; to construct the triangle.

Let AB be equal to the given segment intercepted between the perpendicular and the vertical angle; BC equal to the given difference of the sides ; draw BD perpendicular to AB, and make BE=BA; join AE, and

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produce it; and from C to AE draw CF such that its square may be equal to the given squares of the base and segment AB; draw FG perpendicular to AB; and make GH=BC. From H to BD, draw HD equal to the given base ; join AD; AHD is the triangle required.

For AB is made equal to the given segment, and HD to the given base. Also since AB=BE, AG=GF, and HB=GC; whence the squares of GA and HB are equal to the squares of GA and GC, i. e. to the square of CF, or the squares of HD and AB, by construction. But (iv. 29.) the squares of AD and HB are equal to the squares of HD and AB; .. the squares of GA and HB are equal to the squares of AD and HB; whence GA= AD, and ... the difference between AH and AD is equal to HG, i.e. to BC, or the given difference.

(46.) Given the vertical angle, the side of the in

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square, and the rectangle contained by one side and its segment adjacent to the base made by the angular point of the inscribed square. To construct the triangle.

Let AB be equal to a side of the inscribed square ;

and upon it describe a segment of a circle containing an angle equal to the given angle. From A draw AC perpendicular and equal to AB; and through C draw DCE parallel to AB. Find the centre of the circle; and from it to DE, draw OD such that the difference of the squares of OD

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and the radius of the circle may be equal to the given rectangle. Join DA, and produce it to F. Join FB, and produce it to E; DFE is the triangle required.

From D draw the tangent DG. Join GO. Then the squares of DG, GO are together equal to the square of DO, i.e. to the square of GO and the given rectangle; and .; the square of DG, i.e. the rectangle AD, DF is equal to the given rectangle; .. DF is a side of the triangle, in which AB is the side of an inscribed square. And DFE is equal to the given vertical angle.

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