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WOLFRAM|DEMONSTRATIONS PROJECT

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
.
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