The Plemelj Construction of a Triangle: 2
The Plemelj Construction of a Triangle: 2
This Demonstration constructs a triangle given the length of its base, the length of the altitude from to and the difference between the angles at and .
ABC
AB=c
h
C
C
AB
α-β
A
B
Construction
Step 1: Draw a line segment of length . Draw the line segment of length perpendicular to . Let be the midpoint of .
AB
c
AA'
2
h
C
AB
E
A'B
Step 2: Construct a circle such that the chord subtends an angle from points on above the chord. Let be the center of . The corresponding central angle .
σ
A'B
δ
σ
S
σ
∠A'SB=2δ
Step 3: Let be the intersection of and the line through parallel to .
C
σ
E
AB
Step 4: The triangle meets the stated conditions.
ABC
Verification
Let be the midpoint of . By construction, , so .
D
AB
AA'=2
h
C
DE=
h
C
By construction, , and the altitude from to has length .
AB=c
C
AB
h
C
It remains to prove .
δ=α-β
Let the angle at be . Consider these three angles around the point : , and , which sum to . Since is parallel to and is the midpoint of , is the perpendicular bisector of , is isosceles, bisects and alternate angles around imply . Therefore . Since , . (1)
C
γ
C
∠ACA'
∠A'CB
∠ACB=γ
2π
CE
AB
E
A'B
CE
AA'
ACA'
CE
∠ACA'
AC
∠ACA'=2α
∠A'CB=2π-2α-γ
α+β+γ=π
∠A'CB=2π-2α-(π-α-β)=π+β-α
On the other hand, the chord subtends the angle from any point on above : . The quadrilateral is cyclic, so its opposite angles are supplementary: or . (2)
A'B
δ
F
σ
AB
∠A'BF=δ
A'CBF
∠A'CB+∠A'BF=π
∠A'CB=π-δ
From (1) and (2), , so .
π+β-α=π-δ
δ=α-β