For AB is the given base; BE is the given line; and AE : EC :: AD: DB, i. e. in the given ratio; and the angle at C is equal to AED, i. e. to the given angle.

(12.) Given the perpendicular, the line bisecting the vertical angle, and the line bisecting the base; to construct the triangle.

H

E FC L B

G

From any point C in the indefinite line AB, draw a perpendicular CD equal to the given perpendicular; and with D as centre, and radii equal to the two given lines describe circles cutting AB in E and F. Through E draw GEH perpendicular to AB; join DE, DF; and produce DF to meet HE in G. Bisect DG in I; and draw IO at right angles to DG, meeting GH in O. With the centre O, and radius OG describe a circle cutting AB in K and L; join DK, DL; DKL is the triangle required.

Join OD, OK, OL. Since OI bisects DG at right angles, OD=OG, and the circle passes through D. And since OE is perpendicular to KL, KE=EL, or KL is bisected by DE, which is equal to the given bisecting line; and the arc KG = GL, and the angle KDF is equal to FDL, or the angle KDL is bisected by DF, which is equal to the given line; and DC was made equal to the given perpendicular.

(13.) Given the line bisecting the vertical angle,

PP

the line bisecting the base, and the difference of the angles at the base; to construct the triangle.

Construct a right-angled triangle FDC, (see last Fig.) having its hypothenuse FD equal to the given line which bisects the vertical angle, and the angle FDC equal to half the given difference of the angles at the base. Produce FC both ways; and to it from D draw DE equal to the given line which bisects the base. Draw HEG parallel to DC, meeting DF produced in G. Bisect GD in I; and from I draw 10 at right angles to DG, meeting GH in O. With the centre O, and radius OG, describe a circle, cutting FC produced in K and L; join DK, DL; DKL is the triangle required.

For KE=EL, i. e. the base KL is bisected by DE, which is equal the given line; and the angle KDF is equal to FDL, being on equal circumferences KD, DL; i. e. the vertical angle KDL is bisected by DF, which was made equal to the given bisecting line. Also (iii. 5.) the difference between the angles DLK and DKL is equal to twice the angle FDC, i. e. to the given angle.

(14.) Given the vertical angle, and the line drawn to the base bisecting the angle and the difference between the base and the sum of the sides; to construct the triangle.

Let ABC be equal to the given angle, and BD the line bisecting it. Make BE and BF, each equal to half the given difference. From F draw FO perpendicular to BC, meet

E

D G

B

ing BD in O. With the centre O, and radius OF describe a circle, it will touch AB in E (Eucl. iv. 4.). Through D draw a line AC touching the circle in G; ABC will be the triangle required.

For (Eucl. iii. 36. Cor.) AE=AG, and GC=CF, .. AC is equal to AE and CF together; whence the difference between AC and the sum of the sides AB, BC is equal to BE, BF together, i. e. to the given difference. Also BD is equal to the given line, and it bisects the angle ABC, which is equal to the given vertical angle.

(15.) Given the line bisecting the vertical angle, the perpendicular drawn to it from one of the angles at the base, and the other angle at the base; to construct the triangle.

"Let AB be equal to the given bisecting line; and upon it describe a segment of a circle containing an angle equal to the given angle. Draw BC perpendicular to · AB, and make BD equal to the given

F

H

B

E

K

perpendicular. Bisect AB in E; join ED, and produce it to F; join FA, FB; and through D, draw GDH parallel to AB. In FB produced take BI equal to BH. Join AI; AFI is the triangle required.

Join IG, cutting AB in K. Because GH is parallel to AB, and FE bisects AB, it also bisects GH, i. e. GD =DH; but HB also is equal to BI; .. BD is parallel to GI, and IK is half of IG, and.. equal to BD the

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.

G

H

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. GC, CF; GCF is the triangle required.

Join

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.

B

E

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

BE

BC equal to BA; and upon BC describe a semicircle