# 26c. Construct a Triangle Given the Length of Its Base, the Angle Opposite the Base and the Length of That Angle's Bisector

26c. 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 the point , and the length of the angle bisector at . The Demonstration uses the conchoid of Nicomedes, which is shown in red.

ABC

c

AB

γ

C

C

Draw a right-angled triangle with hypotenuse and . Let be on such that . Let and be on the perpendicular to at on either side of such that and .

COD

CO=b

C

∠DOC=γ/2

B

CD

BC=c

F

O

CD

D

D

DF=c

DO=b

C

Define the conchoid determined by and .

DF

DO

Let be the line from that forms an angle with the horizontal line . Let be the foot of the perpendicular from to .

L

C

γ

CD

E

O

L

Let be the intersection of and the conchoid. Let be the intersection of and . Then triangle is the solution.

A

L

B

CD

AO

ABC

By definition of the conchoid determined by the base with pole and the number , . By construction, . Since and are perpendicular projections of of the same size, is the angle bisector at and by construction .

CD

O

c

BA=c

∠BCA=γ

D

E

O

CO

C

CO=b

C