26b. Construct a Triangle Given the Length of Its Base, the Angle Opposite the Base and the Length of That Angle's Bisector
26b. Construct a Triangle Given the Length of Its Base, the Angle Opposite the Base and the Length of That Angle's Bisector
This Demonstration constructs a triangle given the length of its base , the angle at and the length of the angle bisector from to .
ABC
c
AB
γ
C
b
C
C
AB
Construction
1. Draw the segment of length . Draw a circle with center and central angle over the chord .
AB
c
σ
1
S
2γ
AB
2. Let be the diameter of perpendicular to . Let be the midpoint of .
ED
σ
1
AB
M
AB
3. Let the point be on the extension of the segment such that . Draw a circle with center and radius /2.
F
EA
FA=/2
b
C
σ
2
F
b
C
4. Draw the line through and that intersects at the points and .
D
F
σ
2
G
H
5. Draw the circle with center and radius . The point is one of the points of intersection of and . Draw triangle .
σ
3
D
DH
C
σ
1
σ
3
ABC
6. Let be the intersection of and . Let be the circumcircle of triangle with center .
K
AB
CD
σ
4
AKC
Q
Verification
Theorem: The segment is the angle bisector of .
CK
∠ACB
In the isosceles triangle , and . Since , . So is on . Therefore is a tangent of and the power of with respect to the circle is .
AQK
∠AQK=γ/2
∠QAK=(α+β)/2
∠KAD=γ/2
∠QAD=π/2
Q
AE
AD
σ
4
D
DCDK=
2
AD
The power of with respect to is . By construction . So , .
D
σ
2
DHDG=
2
AD
DH=DC
DG=KD
CK=HG=
b
C