The Plemelj Construction of a Triangle: 3
The Plemelj Construction of a Triangle: 3
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
c
AB
h
C
C
AB
δ=α-β
A
B
Let .
m=+
2
c
2
(2)
h
C
Construction
Step 1: Draw a right-angled triangle with legs of length and , so that .
ABK
AB=c
BK=2
h
C
AK=m
Step 2: Construct a circle with center through (therefore of radius ). Construct the ray from such that the angle between and is . Let be the intersection of and .
σ
A
K
m
ρ
B
ρ
BK
δ
K'
σ
ρ
Step 3: Let be the midpoint of , so . Let be the midpoint of and let be the intersection of , the right bisector of , with the line through parallel to .
H
BK
BH=
h
C
M
KK'
C
AM
KK'
H
AB
Step 4: The triangle meets the stated conditions.
ABC
Verification
Since , and . The line is the right bisector of , so . So .
AB||CH
∠MCH=α
∠BCH=β
CH
KB
∠KCH=∠BCH=β
∠MCK=∠MCH-∠KCH=α-β
Now , so . is isosceles, so also. Then . Therefore in the isosceles triangle , the equal angles at and are . On the other hand, from the right angle at , . So , .
α=∠MCH=∠KCH+∠KCM=β+∠KCM
∠KCM=α-β
CKK'
∠K'CM=α-β
∠BCK'=∠BCH+∠HCK+∠KCM+∠MCK'=β+β+(α-β)+(α-β)=2α
BCK'
B
K'
∠CBK'=∠BK'C=π/2-α
B
∠CBK'=π/2-β-δ
α=β+δ
δ=α-β