# The Plemelj Construction of a Triangle: 2

The Plemelj Construction of a Triangle: 2

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 line segment of length . Draw the line segment of length perpendicular to . Let be the midpoint of .

AB

c

AA'

2

h

C

AB

E

A'B

Step 2: Construct a circle such that the chord subtends an angle from points on above the chord. Let be the center of . The corresponding central angle .

σ

A'B

δ

σ

S

σ

∠A'SB=2δ

Step 3: Let be the intersection of and the line through parallel to .

C

σ

E

AB

Step 4: The triangle meets the stated conditions.

ABC

Verification

Let be the midpoint of . By construction, , so .

D

AB

AA'=2

h

C

DE=

h

C

By construction, , and the altitude from to has length .

AB=c

C

AB

h

C

It remains to prove .

δ=α-β

Let the angle at be . Consider these three angles around the point : , and , which sum to . Since is parallel to and is the midpoint of , is the perpendicular bisector of , is isosceles, bisects and alternate angles around imply . Therefore . Since , . (1)

C

γ

C

∠ACA'

∠A'CB

∠ACB=γ

2π

CE

AB

E

A'B

CE

AA'

ACA'

CE

∠ACA'

AC

∠ACA'=2α

∠A'CB=2π-2α-γ

α+β+γ=π

∠A'CB=2π-2α-(π-α-β)=π+β-α

On the other hand, the chord subtends the angle from any point on above : . The quadrilateral is cyclic, so its opposite angles are supplementary: or . (2)

A'B

δ

F

σ

AB

∠A'BF=δ

A'CBF

∠A'CB+∠A'BF=π

∠A'CB=π-δ

From (1) and (2), , so .

π+β-α=π-δ

δ=α-β