The Plemelj Construction of a Triangle: 7
The Plemelj Construction of a Triangle: 7
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 . This is not Plemelj's construction, but a new one that unifies Plemelj's first construction and an alternative one.
ABC
AB=c
h
C
C
AB
α-β
A
B
Construction
Step 1: Draw a straight line of length and a perpendicular line segment with midpoint .
AB
c
AA'
H
Step 2: Draw a circle with center such that is viewed at an angle from points on below the chord . Let be the midpoint of . The angle equals .
σ
S
BA'
π/2-δ
σ
A'B
G
A'B
SBG
δ
Step 3: Find a point on the circle at distance from and a point at distance from .
B'
c
A'
K
c
B
Step 4: Draw the isosceles trapezoid .
B'BKA'
Step 5: The point is the intersection of the straight line through parallel to and the right bisector of and .
C
H
AB
BB'
KA'
Step 6: The triangle meets the stated conditions.
ABC
Verification
This is similar to Plemelj's first construction, but instead of triangle , start with triangle , which is also congruent to . In the isosceles triangle , , so . The obtuse angle ; ∠A'CK=1/2(π-(α-β)-γ)=β.
A'B'C
BCK
ABC
B'BC
∠CBB'=π/2-δ-β
∠B'CB=2(δ+β)
∠A'CB=π-(α-β)
1
2
On the other hand, ∠A'CK also equals . Thus .
1
2
1/2(2π-2(δ+β)-2γ)=α-δ
δ=α-β