# 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