# The Plemelj Construction of a Triangle: 10

The Plemelj Construction of a Triangle: 10

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 is an alternative to Plemelj's second construction. See The Plemelj Construction of a Triangle: 3.

ABC

c

AB

h

C

C

AB

δ

α

A

β

B

Let .

m=+

2

c

2

(2)

h

C

Construction

Draw a right-angled triangle with legs of length and , so that .

ABK

AB=c

AK=2

h

C

BK=m

Step 1: Construct a circle with center through and, therefore, of radius . Construct the ray from such that the angle between and is . Let be the intersection of and .

σ

B

K

m

ρ

A

ρ

AK

δ

K'

σ

ρ

Step 2: Let be the midpoint of , so that . Let be the midpoint of . Let be the intersection of the perpendicular bisector of and the line through parallel to .

H

AK

AH=

h

C

M

KK'

C

KK'

H

AB

Step 3: The triangle meets the stated conditions.

ABC

Verification

Since and is the perpendicular bisector of , and . So . The triangle is isosceles, so also and , and .

AB||HC

MB

KK'

∠MCH=β

∠ACH=α

∠MCK=∠KCH-∠HCM=α-β

CKK'

∠K'CM=α-β

∠K'CK=2(α-β)

∠ACK'=2α-2(α-β)=2β

Since is isosceles, .

K'AC

∠ACK'=π-2(π/2-α+δ)=2(α-δ)

So and .

2β=2(α-δ)

δ=α-β