26c. Construct a Triangle Given the Length of Its Base, the Angle Opposite the Base and the Length of That Angle's Bisector

γ

0.8

b

C

1

c

1.5

plot range

2.4

This Demonstration constructs a triangle

ABC

given the length

c

of its base

AB

, the angle

γ

at the point

C

, and the length of the angle bisector at

C

. The Demonstration uses the conchoid of Nicomedes, which is shown in red.

Draw a right-angled triangle

COD

with hypotenuse

CO=

b

C

and

∠DOC=γ/2

. Let

B

be on

CD

such that

BC=c

. Let

F

and

O

be on the perpendicular to

CD

at

D

on either side of

D

such that

DF=c

and

DO=

b

C

.

Define the conchoid determined by

DF

and

DO

.

Let

L

be the line from

C

that forms an angle

γ

with the horizontal line

CD

. Let

E

be the foot of the perpendicular from

O

to

L

.

Let

A

be the intersection of

L

and the conchoid. Let

B

be the intersection of

CD

and

AO

. Then triangle

ABC

is the solution.

By definition of the conchoid determined by the base

CD

with pole

O

and the number

c

,

BA=c

. By construction,

∠BCA=γ

. Since

D

and

E

are perpendicular projections of

O

of the same size,

CO

is the angle bisector at

C

and by construction

CO=

b

C

.