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CO to F, making OF=0A; join AF, FB; AFB is the triangle required.

Join OB. Because AD=DB, and DO is common, and the angles at D right angles, .. OB=0A=OF; and a circle described from O as a centre, and radius OA, will pass through F and B, and circumscribe the triangle ABF. And FOC produced is a diameter. Also the angles AOC, AOF are equal to two right angles, i.e. to AOC and twice the given angle; .. AOF is equal to twice the given angle ; and since it is double of ABF, ABF will be equal to the given angle.

(30.) Given the vertical angle, the base, and the difference between two lines drawn from the centre of the inscribed circle to the angles at the base ; to construct the triangle.

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Take any line AB, unlimited towards B, and cut off AC equal to the given difference; and at the point C make the angle BCD such that its quadruple together with the given angle at the vertex may be equal to two right angles, From A to CD, draw AD equal to the given base.

At the point A make the angle BAE=BAD, and at the point D make the angle CDB=DCB, and BDE= BDA; EDA is the triangle required.

For the angles EAD, EDA being bisected by AB, BD, B is the centre of the inscribed circle. And the angle BCD being equal BDC, BC is equal to BD, and ::. the difference between BA and BD is equal to AC, i. e. to the given difference. Also the angles EAD, EDA are together double of the angles BAD, BDA, i. e. of BCD and BCD, and .. are equal to 4 BCD ); whence the angle AED together with 4 BCD will be equal to two right angles ; and :. the angle AED is equal to the given vertical angle. Also AD is equal to the given base.

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(31.) Given that segment of the line bisecting the vertical angle which is intercepted by perpendiculars let fall upon it from the angles at the base ; the ratio of the sides ; and the ratio of the radius of the inscribed circle to the segment of the base which is intercepted between the line bisecting the vertical angle and the point of contact of the inscribed circle ; to construct the triangle.

Let AB be equal to the given segment of the bisecting line; from A and B, on opposite sides, draw AC, BD at right angles to it. On AB, as hypothenuse, describe a right-angled triangle AEB, whose sides are in the given ratio of the radius of the inscribed circle to the segment of the base intercepted between the bisecting line and the point of contact of the inscribed circle. Divide AB in F in the ratio of the sides; and through F, draw CFD parallel to AE, and meeting the perpendiculars in C and D. - Make BG=BD. Join CG, and produce it to meet AB produced in H; join HD; HCD is the triangle required.

Draw CO bisecting the angle HCD. Since BG= BD, and BH is perpendicular to GD, it bisects the

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angle GHD; whence O is the centre of the inscribed circle. From O draw OI perpendicular to CD, and .. parallel to BE; whence

01 : IF :: BE : EA. i. e. in the given ratio of the radius of the inscribed circle to the segment of the base intercepted between the bisecting line and the point of contact of the inscribed circle. Also from the similar triangles AFC, BFD,

AF : FB :: CF : FD :: CH : HD,
:. CH : HD in the given ratio of the sides.

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(32.) Given the line bisecting the vertical angle, and the differences between each side and the adjacent segment of the base made by the bisecting line ; to construct the triangle.

Let AB be equal to the given bisecting line. Produce it to C, so that. BC may be a fourth proportional to AB and the given differences. On AC, as diameter, describe a circle ADE; in which place a line DBE, passing through B, equal to the sum of the given differences ; and from A draw AF, AG to the circumference, each equal to DE; and produce them to meet DE produced in H and I; AHI is the triangle required.

Since DE is equal to the sum of the given differences, and the rectangle DB, BE, is equal to the rectangle AB, BC, i.e. to the rectangle contained by the given differences, DB and BE will be the given differences. And since AF= AG, the arcs AF, AG are equal and AC being a diameter, FC and CG will also be equal, .. the angles FAC, CAG are equal, or HAI is bisected by AC. Also since FA=DE, AH and HE will be equal; .. the difference between AH and HB is equal to BE one of the given differences. In the same manner, AI= ID; and .:. the difference between Al and IB is equal to BD, the other given difference.

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(33.) Given one of the angles at the base, the side opposite to it, and the rectangle contained by the base and that segment of it made by the perpendicular which is adjacent to the given angle ; to construct the triangle.

Let AB be equal to the given side. Upon it describe a segment of a circle containing an angle equal to the given angle. Bisect AB in C; and from C draw to the circumference, a line CD such that the difference between the squares of CD and CB may be equal to the given rectangle. Join AD, DB; ADB is the triangle required.

On AB describe a circle ABE. Join BE. Then the rectangle AD, DE is equal to the rectangle GD, DF, i.e. to the difference of the squares of CD and CG or to the given rectangle. · And BE is perpendicular to AD; AB is equal to the given side, and ADB to the given angle.

(34.) Given the vertical angle, and the lengths of two lines drawn from the extremities of the base to the points of bisection of the sides ; to construct the triangle.

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Let AB be equal to one of the given lines ; bisect it in C; and on AC describe a segment of a circle containing an angle equal to the given angle. From BA cut off BD equal to one third of BA; and from D to the circle, draw DE equal to one third of the other given line. Produce ED, and make DF equal to twice DE. Join AE, FB; and produce them to meet in G. Join AF; AFG is the triangle required.

Join CE. Since BD is double of DC, and DF double of DE; EC is parallel to FG ; and .. the angle AGF is equal to AEC, i. e. to the given angle. Hence also AE is equal to EG; and BF being double of EC, is equal to BG; :. AB and FE, equal to the given lines, are drawn to the points of bisection of the sides.

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(35.) Given the lengths of three lines drawn from the angles to the points of bisection of the opposite sides ; to construct the triangle.

Describe a triangle ABC whose sides are respectively equal to two thirds of the given lines ; and complete the parallelogram ABDC. Join AD; and produce CB to E, making BE=BC. Join AE, ED; AED is the triangle required.

Produce AB, DB to Fand G. Since the diagonals of parallelograms bisect each other, AH is equal to HD, and BH to HC; .. EH is equal to EB and BH together, i. e. to BC and BH or to BC and .:, is equal to

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