The Plemelj Construction of a Triangle: 11

c

h

C

δ = α-β

0.7

steps

1

2

3

4

verification

plot range

1

Plemelj's first construction

This Demonstration constructs a triangle

ABC

given the length

c

of its base

AB

, the length

h

C

of the altitude from

C

to

AB

and the difference

δ

between the angles

α

at

A

and

β

at

B

. This construction is an alternative to the one from Plemelj's teacher's textbook. See The Plemelj Construction of a Triangle: 2.

Construction

Step 1: Draw a line segment

AB

of length

c

. Draw the line segment

BB'

of length

2

h

C

perpendicular to

AB

. Let

E

be the midpoint of

A'B

.

Step 2: Construct a circle

σ

such that the chord

AB'

subtends an angle

π-δ

from points on

σ

above the chord. Let

S

be the center of

σ

. The corresponding central angle is

∠ASB'=2(π-δ)

.

Step 3: Let

C

be the intersection of

σ

and the line through

E

parallel to

AB

.

Step 4: The triangle

ABC

meets the stated conditions.

Verification

By construction,

AB=c

.

Let

D

be the midpoint of

AB

. By construction,

BB'=2

h

C

, so

DE=

h

C

and the altitude from

C

to

AB

has length

h

C

.

It remains to prove

δ=α-β

.

The angle

∠B'CA=γ+2β=π-(α-β)

, since

CEAB

and triangle

BCB'

is isosceles.

By construction, the chord

AB'

subtends the angle

π-δ

from any point above the chord.

So

δ=α-β

.