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given perpendicular. Also since AB bisects GI at right angles, it also bisects the angle IAG; and it is equal to the given bisecting line. And AFI is equal to the other given angle.
(16.) Given the straight line bisecting the vertical angle, and the perpendiculars drawn to that line from the extremities of the base ; to construct the triangle.
On the indefinite straight line AB, take AC, CD respectively equal to the greater and less perpendiculars; and from C draw CE at right angles to AB, and equal to the given bisecting line. Take AB : BD :: AC : CD. Join BE; and produce it to meet AF, DG, drawn from A and D, perpendicular to AB. Join GC, CF; GCF is the triangle required.
Draw FI, GH perpendicular to CE. Then the triangles BGD, ABF being similar,
AF : GD :: AB : BD :: AC : CD, and the angle at A is equal to GDC; whence (Eucl. vi. 6.) AFC, GDC are similar, and the angle ACF is equal to GCD; and .:. the angle FCE is equal to ECG, i. e. the angle FCG is bisected by EC, which was made equal to the given bisecting line. Also FI, GH are respectively equal to AC, CD, which were made equal to the given perpendiculars.
(17.) Given the vertical angle, the difference of the two sides containing it, and the difference of the
segments of the base made by a perpendicular from the vertex; to construct the triangle.
Let AB be equal to the given difference of the segments of the base. Draw BD, making the angle ABD equal to half the given angle; and from A draw AD meeting it in D, and equal to the given difference of the sides ; produce it, and make the angle DBE= EDB, and with the centre E and radius EB describe the circle DBC meeting AB, AD produced in C and F; join EC: AEC is the triangle required.
Join FC. Since BDFC is a quadrilateral figure inscribed in a circle (Eucl. iii. 22.) the angles ABD, DFC are equal; but AEC is double of DFC (Eucl. iii. 20.), and .. also of ABD, i.e. it is equal to the given angle. Also since the angles EDB, EBD are equal, ED=EB = EC, :. AD is the difference of the sides AE, EC, which .. is equal to the given difference; and AB is evidently equal to the difference of the segments of the base.
(18.) Given the base and vertical angle, to construct the triangle, when the square of one side is equal to the square of the base, and three times the square of the other side.
Let AB be equal to the given base. Upon it describe a segment of a circle containing an angle equal to the given angle. Produce AB to C, and make BC equal to BA; and upon BC describe a semicircle
BDC cutting the segment in D. Join AD, BD; ABD is the triangle required.
Let fall the perpendicular DE; then (Eucl. vi. the square of DB is equal to the rectangle CB, BE or to the rectangle AB, BE; and (Eucl. ii. 12.) the square of AD is equal to the squares of AB, BD, and twice the rectangle AB, BE, i. e. to the square of AB and three times the square of BD. Also, by construction, ADB is equal to the given angle.
(19.) Given the base and perpendicular ; to construct the triangle, when the rectangle contained by the sides is equal to twice the rectangle contained by the segments of the base made by the line bisecting the vertical angle.
Let AB be equal to the given base ; and draw the indefinite line ED bisecting it at right angles. Take CE, CD each equal to the given perpendicular; and through the points A, D, B describe a circle. Draw EF parallel to AB meeting the circle in F. Join AF, FB; AFB is the triangle required.
Draw FG perpendicular to AB, it is equal to the given perpendicular. Join FD. Since DC is equal to CE, DF is equal to twice DH; and .:the rectangle DF, FH is double of the rectangle DH, HF, i.e. it is double of the rectangle AH, HB, contained by the segments of the base, made by DF which bisects the angle AFB. And AB was made equal to the given base.
(20.) In a right-angled triangle, having given the sum of the base and hypothenuse, and the sum of the base and perpendicular; to construct the triangle.
Let BD be taken equal to the sum of the base and hypothenuse, and BG equal to the sum of the base and perpendicular. At the point G in the straight line GB, make the angle BGC equal to half a right angle; and let CG be produced to meet a perpendicular to BD drawn through D, in F. Join BF; and from G to BF draw GH=GD. From B draw BC parallel to GH, meeting GC in C; and let fall the perpendicular CA; ACB is the triangle required.
Draw CE parallel to AD. Because GH is equal to GD, BC is equal to CE, į e. to AD; and :: BC and BA together are equal to BD, the given sum of the hypothenuse and base. And since AGC is half a right angle, and the angles at A right angles, AG is equal to AC, :. BA and AC together are equal to BG, the given sum of the base and perpendicular.
(21.) Given the perimeter of a right-angled triangle whose sides are in geometrical progression ; to construct the triangle.
Let AB be equal to the given perimeter; and on it describe a semicircle ACB. Divide AB in extreme and mean ratio in D; and from D draw DC at
right angles to AB. Join AC, CB. Bisect the angles CAB, CBA by the lines AE, BE, meeting in E; and draw EF, EG respectively parallel to AC, CB; FEG is the triangle required.
Since FE is parallel to AC, the angle FEA=EAC= E AF, and .. AF=FE. And in the same manner it may be shewn that EG=GB. Hence GF, FE, EG are together equal to AB the given perimeter. And the angles at F and G being equal to the angles BAC, BCA, the angle FEG is equal to ACB, and is :. a right angle.
Also since AB : AD :: AD : DB and (Eucl. vi. 8. Cor.) AB : BC :: BC : DB,
.::AD=BC; whence also AB : AC :: AC : (AD=) BC; and since the triangles ABC, FGE are equiangular,
FG : FE :: FE : EG, i.e. the sides are in geometrical progression.
(22.) Given the difference of the angles at the base, the ratio of the segments of the base made by the perpendicular, and the sum of the sides ; to construct the triangle. Take
any line AB, and divide it in C, in the given ratio of the segments. On AB describe a segment of a circle containing an angle equal to the given difference. From C draw the perpendicular CD. Join AD, DB; and take DE : DA in the ratio of the given sum, to the sum of AD, DB; and through E draw EF parallel to AB; and make the angle GDH equal to GDF, and