# The Plemelj Construction of a Triangle: 11

The Plemelj Construction of a Triangle: 11

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 at . This construction is an alternative to the one from Plemelj's teacher's textbook. See The Plemelj Construction of a Triangle: 2.

ABC

c

AB

h

C

C

AB

δ

α

A

β

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

BB'

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 is .

σ

AB'

π-δ

σ

S

σ

∠ASB'=2(π-δ)

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

By construction, .

AB=c

Let be the midpoint of . By construction, , so and the altitude from to has length .

D

AB

BB'=2

h

C

DE=

h

C

C

AB

h

C

It remains to prove .

δ=α-β

The angle , since and triangle is isosceles.

∠B'CA=γ+2β=π-(α-β)

CEAB

BCB'

By construction, the chord subtends the angle from any point above the chord.

AB'

π-δ

So .

δ=α-β