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

.