# 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

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 given the angle at , the length of the angle bisector at and the length of the perpendicular from to the bisector.

ABC

γ

C

χ

A

A

p

B

Construction

Step 1: Draw the line segment of length and the line segment of length perpendicular to at . Extend to so that as well.

AD

χ

A

DG

p

AD

D

DG

H

GH=p

Step 2: Draw a circle with center subtending the central angle over the chord so that .

S

2γ

AH

∠SAH=∠SHA=π/2-γ

Let the point be the intersection of the circle and the line through parallel to .

I

G

AD

Step 3: Let be the point on such that is the midpoint of . Construct the point on the ray so that .

J

IG

G

IJ

B

JD

JD=DB

Step 4: Let be the intersection of and .

C

AI

BD

Then satisfies the conditions.

ABC

Verification

Let be the intersection of and .

L

AD

BI

The distance of to is . The segment bisects the angle at since it is the right bisector of .

B

AD

BL=LI=DG=p

AD

A

IB

Since the triangles and are congruent, the quadrilateral is a rhombus. Since , , the angle at is because the lines and are parallel.

IGH

JGD

IDJH

∠AIH=π-γ

∠HIC=γ

C

∠ACB=∠ICB=γ

IH

CD