WOLFRAM|DEMONSTRATIONS PROJECT

The Plemelj Construction of a Triangle: 2

​
c
h
C
δ = α-β
0.7
steps
1
2
3
4
verification
plot range
1
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
.
Construction
Step 1: Draw a line segment
AB
of length
c
. Draw the line segment
AA'
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
A'B
subtends an angle
δ
from points on
σ
above the chord. Let
S
be the center of
σ
. The corresponding central angle
∠A'SB=2δ
.
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
Let
D
be the midpoint of
AB
. By construction,
AA'=2
h
C
, so
DE=
h
C
.
By construction,
AB=c
, and the altitude from
C
to
AB
has length
h
C
.
It remains to prove
δ=α-β
.
Let the angle at
C
be
γ
. Consider these three angles around the point
C
:
∠ACA'
,
∠A'CB
and
∠ACB=γ
, which sum to
2π
. Since
CE
is parallel to
AB
and
E
is the midpoint of
A'B
,
CE
is the perpendicular bisector of
AA'
,
ACA'
is isosceles,
CE
bisects
∠ACA'
and alternate angles around
AC
imply
∠ACA'=2α
. Therefore
∠A'CB=2π-2α-γ
. Since
α+β+γ=π
,
∠A'CB=2π-2α-(π-α-β)=π+β-α
. (1)
On the other hand, the chord
A'B
subtends the angle
δ
from any point
F
on
σ
above
AB
:
∠A'BF=δ
. The quadrilateral
A'CBF
is cyclic, so its opposite angles are supplementary:
∠A'CB+∠A'BF=π
or
∠A'CB=π-δ
. (2)
From (1) and (2),
π+β-α=π-δ
, so
δ=α-β
.