WOLFRAM|DEMONSTRATIONS PROJECT

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

​
c
1.15
γ
1
b
C
0.6
plot range
steps
1
2
3
4
5
6
This Demonstration constructs a triangle
ABC
given the length
c
of its base
AB
, the angle
γ
at
C
and the length of the angle bisector
b
C
from
C
to
AB
.
Construction
1. Draw the segment
AB
of length
c
. Draw a circle
σ
1
with center
S
and central angle
2γ
over the chord
AB
.
2. Let
ED
be the diameter of
σ
1
perpendicular to
AB
. Let
M
be the midpoint of
AB
.
3. Let the point
F
be on the extension of the segment
EA
such that
FA=
b
C
/2
. Draw a circle
σ
2
with center
F
and radius
b
C
/2
.
4. Draw the line through
D
and
F
that intersects
σ
2
at the points
G
and
H
.
5. Draw the circle
σ
3
with center
D
and radius
DH
. The point
C
is one of the points of intersection of
σ
1
and
σ
3
. Draw triangle
ABC
.
6. Let
K
be the intersection of
AB
and
CD
. Let
σ
4
be the circumcircle of triangle
AKC
with center
Q
.
Verification
Theorem: The segment
CK
is the angle bisector of
∠ACB
.
In the isosceles triangle
AQK
,
∠AQK=γ/2
and
∠QAK=(α+β)/2
. Since
∠KAD=γ/2
,
∠QAD=π/2
. So
Q
is on
AE
. Therefore
AD
is a tangent of
σ
4
and the power of
D
with respect to the circle is
DCDK=
2
AD
.
The power of
D
with respect to
σ
2
is
DHDG=
2
AD
. By construction
DH=DC
. So
DG=KD
,
CK=HG=
b
C
.