25. Construct a Triangle Given Its Base, the Difference of the Base Angles and the Length of One of Three Line Segments
25. Construct a Triangle Given Its Base, the Difference of the Base Angles and the Length of One of Three Line Segments
This Demonstration constructs a triangle given the length of its base , the difference between the angles and of the base and the length of one of the following:
ABC
c
AB
δ
α
β
i. the altitude from to the base
C
ii. the bisector from to the base
C
iii. the line segment through the circumcenter from to the base
C
The last two cases are reduced to the first case, since the angle between the bisector and the altitude is , and the angle between the altitude and the line through the center is . So we have a Plemelj triangle, for which we use The Plemelj Construction of a Triangle: 13.
(α-β)/2
α-β
Construction
1. Draw line of length .
AB
c
2. Draw a circle of radius with center , the midpoint of .
σ
1
c/2
M
AB
3. Draw a ray from at an angle to intersect at the point . Let be a point such that the length of is the length of the given altitude and is perpendicular to .
A
δ
σ
1
P
N
MN
MN
AB
4. Draw a circle with center and radius . Let the point be on the circle such that . The point is the intersection of the perpendicular bisector of and the line parallel to through .
σ
2
N
AN
D
AP=AD
C
PD
AB
N
Verification
The angles and are equal to . The angle at is .
∠DAC
∠CAP
β
A
α=β+δ
The first case was posed to Josip Plemelj (1873–1967) by his mathematics teacher in secondary school in 1891 (in Ljubljana, then in the Austro-Hungarian Empire, now in Slovenia).