5. Construct a Triangle Given Its Circumradius, the Difference of Base Angles and the Sum of the Other Two Sides

This Demonstration constructs a triangle

ABC

given the circumradius

R

, the difference of angles

δ

at the base

AB

and the sum

s

of the lengths of the sides

BC

and

CA

.

Construction

Draw a (blue) circle

σ

1

with center

S

, radius

R

and a diameter

DE

. Let

C

be a point on

σ

1

such that

∠CED=δ/2

.

Step 1: Draw a (red) circle

σ

2

with diameter

CE

and center

M

, the midpoint of

CE

. Let

F

be on

σ

2

such that

CF=s/2

.

Step 2: Let

CF

intersect

σ

1

at

B

.

Step 3: Let

A

be symmetric to

B

across

DE

.

Step 4: Draw the triangle

ABC

.

Verification

Let

BC=a

and

CA=b

.

Let

G

be symmetric to

A

across

CE

. Let

P

be the intersection of

AB

and

CE

.

The angle between the angle bisector from

C

and the altitude from

C

is

δ/2

; also,

∠SCM=δ/2

.

So

CM

is the bisector of the angle at

C

. Since

CA=CG=b

,

CF=s/2

and

FG=FB

,

CB=a

, so

s=a+b

.