to CD, i. e. to the given difference. Also the angle at A is equal to the difference between the angles CDB, DBA, or CDB, BDA, i. e. to the given angle. And CB was made equal to the given side.

(3.) Given the base and one of the angles at the base; to construct the triangle, when the side opposite the given angle is equal to half the sum of the other side and a given line.

E

Let AB be the given base, and ABC the given angle; produce CB to D, making BD equal to the given line. Join AD; and from B to AD draw BE, equal to half BD. From A draw AC parallel to BE; ABC is the triangle required. For AB and ABC are made equal to the given base and angle; and since BE is parallel to AC,

AC CD: BE : BD :: 1 : 2.

(4.) Given the base of a right-angled triangle, and the sum of the hypothenuse and a straight line, to which the perpendicular has a given ratio; to construct the triangle.

Let AB be equal to the given base. From B draw BC perpendicular to it, and such that it may have to the given sum, the given

ratio. Join CA, and produce it; and from B to CD, draw BD equal to the given sum. From A draw AE perpendicular to AB; ABE is the triangle required. For AE being parallel to BC,

AE ED: BC: BD, i. e. in the given ratio; .. AE is equal to the perpendicular; and AB was made equal to the given base.

(5.) Given the perpendicular drawn from the vertical angle to the base, and the difference between each side and the adjacent segment of the base made by the perpendicular; to construct the triangle,

D

AIC FB

From any point C in an indefinite line AB, erect a perpendicular CD equal to the given perpendicular; and take CE equal to the given difference between the side and adjacent segment on the opposite side of the perpendicular. Also take CF equal to the other given difference. Join ED, FD; and make the angle FDA=DFA, and EDB=DEB; ADB is the triangle required.

Since the angles ADF, AFD are equal, AD=AF, ..the difference between AD, AC is equal to CF, i. e. to the given difference. In the same manner the difference between BD and BC is equal to CE, i. e. to the given difference.

(6.) Given the vertical angle, and the base; to construct the triangle, when the line drawn from the vertex cutting the base in any given ratio, bisects the vertical angle.

B

Let AB be equal to the given base; and upon it describe a segment of a circle containing an angle equal to the given angle; and let the base be divided in the given ratio in D. Complete the circle; and bisect the arc ACB in C; join CD, and produce it to E; join AE, EB; AEB is the triangle required.

For AEB is equal to the given angle; and since the arc AC=CB, the line ED, which divides AB in the given ratio in D, makes the angle AED=DEB.

(7.) Given the vertical angle, and one of the sides containing it; to construct the triangle, when the line drawn from the vertex making a given angle with the base, bisects the triangle.

Let AB be equal to the given side; and on it describe a segment of a circle containing an angle equal to the given angle made by the bisecting line with the base; and make the angle ABC equal to the given vertical angle.

Bisect AB in D; and draw DE parallel to BC, meeting the circle in E; join AE, and produce it to C; ABC is the triangle required.

Join BE. Since DE is parallel to BC, (Eucl. vi. 2.) AE is equal to EC, and .. it makes with the base AE

BE bisects the triangle; and an angle equal to the given

angle. Also AB is equal to the given side, and ABC to the given angle at the vertex.

(8.) Given one angle, a side opposite to it, and the sum of the other two sides; to construct the triangle.

D

Let AB be the given side. Upon it describe a segment of a circle containing an angle equal to half the given angle; and from A draw AC equal to the given sum of the two sides; join BC; and make the angle CBD equal to BCD; ABD is the triangle required.

B

Since the angle DCB is equal to DBC, DB is equal to DC; .. AD, DB together are equal to the given sum. And the angle ADB is equal to DBC, DCB, i. e. to twice DCB, and .. is equal to the given angle.

(9.) Given the vertical angle, the line bisecting the base, and the angle which the bisecting line makes with the base; to construct the triangle.

On any line AB describe a segment of a circle containing an angle equal to the given angle. Bisect AB in C; and at C make the angle

E

D

B

BCD equal to the given angle which the bisecting line makes with the base; produce it, if necessary, till CE is equal to the given line; draw EF, EG respectively parallel to DA, DB; EFG is the triangle required.

For FE and EG being respectively parallel to DA, ᎠᏴ,

FC: CE (AC=) CB : CD :: CG : CE, .. FC=CG, and EC, which is equal to the given line, bisects the base; and the angles FEC, CEG being equal

to the angles ADC, CDB, FEG is equal to ADB, i.e. to the given angle.

(10.) Given the vertical angle, the perpendicular drawn from it to the base, and the ratio of the segments of the base made by it; to construct the triangle.

line AB, and on it describe

E

C

Take any a segment of a circle containing an angle equal to the given angle. Divide AB in F C, in the given ratio; and from C draw the perpendicular CD, from which cut off DE equal to the given perpendicular. Join DA, DB; and through E draw FEG parallel to AB; DFG is the triangle required.

Since FG is parallel to AB,

FE: EG :: AC: CB, i. e. in the given ratio, and DE is equal to the given perpendicular, and FDG to the given angle.

(11.) Given the vertical angle, the base, and a line drawn from either of the angles at the base to cut the opposite side in a given ratio; to construct the triangle.

e;

Let AB be equal to the given base and divide it at D in the given ratio. On AD describe a segment of a circle

containing an angle equal to the given angle; and from B draw BE equal to the given line. Join AE, ED; and from B draw BC parallel to DE, and meeting AE produced in C; ABC is the triangle required.