20b. Construct a Triangle Given the Length of Its Base, the Difference of Its Base Angles and a Line Containing the Third Vertex
20b. Construct a Triangle Given the Length of Its Base, the Difference of Its Base Angles and a Line Containing the Third Vertex
This Demonstration constructs a triangle given the length of its base, the difference of the base angles and a line (given by a point and slope ) that contains . This generalizes the problem given to Plemelj when he was in high school, namely, the problem in which the line is parallel to . Our construction is a simple adaptation of The Plemelj Construction of a Triangle: 4.
ABC
c
δ
λ
N
ϵ
C
λ
AB
Construction
Step 1: Draw a line segment of length and its midpoint . Draw a line .
AB
c
M
λ
Step 2: Draw a line segment from that is perpendicular to . It meets the line at .
M
AB
λ
N
Step 3: Choose any point on the ray .
O'
NM
Step 4: Let be on such that .
C'
λ
∠NO'C'=δ
Step 5: Let be the circle with center and radius .
σ
O'
O'C'
Step 6: Let be the intersection of and the ray . Join and .
A'
σ
NA
A'C'
A'O'
Step 7: The point is the intersection of and the line through parallel to .
C
λ
A
A'C'
Step 8: The triangle meets the stated conditions.
ABC
Verification
Theorem: Let be any triangle, and let be the foot of the altitude from to . Let be the circumcenter of , and let the segment be perpendicular to with on the same side of as . Then .
ABC
D
C
AB
O
ABC
ON
AB
N
AB
C
∠DCO=∠CON=α-β
Proof: The inscribed angle subtended by the chord is , so the central angle . Since is isosceles, . Since , . Since , [1, Proposition 29]. So .
AC
β
∠AOC=2β
AOC
∠ACO=π/2-β
∠ACD=π/2-α
∠DCO=∠ACO-∠ACD=π/2-β-(π/2-α)=α-β
DC||ON
∠CON=∠DCO
∠CON=α-β
Now in the construction, let be the intersection of and the line through parallel to . Then the isosceles triangles and are similar, and is parallel to , so . By construction, , so . Since is on the right bisector of , it is the circumcenter of . By the theorem, .
O
MN
A
A'O'
A'O'C'
AOC
O'C'
OC
∠NO'C'=∠NOC
∠NO'C'=δ
∠NOC=δ
O
AB
ABC
δ=α-β