The Plemelj Construction of a Triangle: 7

c

h

C

δ = α-β

0.62

steps

1

2

3

4

5

6

additional labels

characteristic triangles

from first construction

for alternative construction

plot range

0.9

photograph from archive

This Demonstration constructs a triangle

ABC

given the length

AB=c

of its base, the length

h

C

of the altitude from

C

to

AB

and the difference

α-β

between the angles at

A

and

B

. This is not Plemelj's construction, but a new one that unifies Plemelj's first construction and an alternative one.

Construction

Step 1: Draw a straight line

AB

of length

c

and a perpendicular line segment

AA'

with midpoint

H

.

Step 2: Draw a circle

σ

with center

S

such that

BA'

is viewed at an angle

π/2-δ

from points on

σ

below the chord

A'B

. Let

G

be the midpoint of

A'B

. The angle

SBG

equals

δ

.

Step 3: Find a point

B'

on the circle at distance

c

from

A'

and a point

K

at distance

c

from

B

.

Step 4: Draw the isosceles trapezoid

B'BKA'

.

Step 5: The point

C

is the intersection of the straight line through

H

parallel to

AB

and the right bisector of

BB'

and

KA'

.

Step 6: The triangle

ABC

meets the stated conditions.

Verification

This is similar to Plemelj's first construction, but instead of triangle

A'B'C

, start with triangle

BCK

, which is also congruent to

ABC

. In the isosceles triangle

B'BC

,

∠CBB'=π/2-δ-β

, so

∠B'CB=2(δ+β)

. The obtuse angle

∠A'CB=π-(α-β)

;

1

2

∠A'CK=1/2(π-(α-β)-γ)=β

.

On the other hand,

1

2

∠A'CK

also equals

1/2(2π-2(δ+β)-2γ)=α-δ

. Thus

δ=α-β

.