The Plemelj Construction of a Triangle: 6
The Plemelj Construction of a Triangle: 6
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 vertical segment of length . Draw two horizontal rays and from and , respectively. From , draw a ray at angle with respect to the ray . Let be the intersection of the rays and .
DE
h
C
ρ
τ
D
E
D
κ
δ
ρ
C
κ
τ
Step 2: Draw a segment of length perpendicular to the ray .
CF
c/2
κ
Step 3: Draw a circle with center and radius . The ray intersects the circle at and .
σ
F
c/2
DF
σ
G
H
Step 4: On , draw the point so that . On , measure out the point so that .
ρ
A
AD=DG
ρ
B
AB=c
Step 5: The triangle satisfies the stated conditions.
ABC
Verification
Draw the circumcircle of the triangle with center . Let the point be the foot of the altitude from . Then . By construction, the power of the point with respect to is =DADB, so by Euclid III.37 [1, pp. 15], is tangent to and as angles with orthogonal legs. So .
σ
2
ABC
S
I
C
∠ICS=α-β
D
σ
2
2
DC
DC
σ
2
δ=∠ICS
δ=α-β