# 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