17b. Construct a Triangle Given the Length of an Angle Bisector, the Angle at Another Vertex and the Distance from the Third Vertex to the Angle Bisector

This Demonstration shows an alternative construction of a triangle

ABC

given the angle

γ

at

C

, the length

χ

A

of the angle bisector at

A

and the length

p

of the perpendicular from

B

to the bisector.

Construction

Step 1: Draw the line segment

AD

of length

χ

A

and the line segment

DG

of length

p

perpendicular to

AD

at

D

. Extend

DG

to

H

so that

GH=p

as well.

Step 2: Draw a circle with center

S

subtending the central angle

2γ

over the chord

AH

so that

∠SAH=∠SHA=π/2-γ

.

Let the point

I

be the intersection of the circle and the line through

G

parallel to

AD

.

Step 3: Let

J

be the point on

IG

such that

G

is the midpoint of

IJ

. Construct the point

B

on the ray

JD

so that

JD=DB

.

Step 4: Let

C

be the intersection of

AI

and

BD

.

Then

ABC

satisfies the conditions.

Verification

Let

L

be the intersection of

AD

and

BI

.

The distance of

B

to

AD

is

BL=LI=DG=p

. The segment

AD

bisects the angle at

A

since it is the right bisector of

IB

.

Since the triangles

IGH

and

JGD

are congruent, the quadrilateral

IDJH

is a rhombus. Since

∠AIH=π-γ

,

∠HIC=γ

, the angle at

C

is

∠ACB=∠ICB=γ

because the lines

IH

and

CD

are parallel.