# The Plemelj Construction of a Triangle: 3

The Plemelj Construction of a Triangle: 3

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

c

AB

h

C

C

AB

δ=α-β

A

B

Let .

m=+

2

c

2

(2)

h

C

Construction

Step 1: Draw a right-angled triangle with legs of length and , so that .

ABK

AB=c

BK=2

h

C

AK=m

Step 2: Construct a circle with center through (therefore of radius ). Construct the ray from such that the angle between and is . Let be the intersection of and .

σ

A

K

m

ρ

B

ρ

BK

δ

K'

σ

ρ

Step 3: Let be the midpoint of , so . Let be the midpoint of and let be the intersection of , the right bisector of , with the line through parallel to .

H

BK

BH=

h

C

M

KK'

C

AM

KK'

H

AB

Step 4: The triangle meets the stated conditions.

ABC

Verification

Since , and . The line is the right bisector of , so . So .

AB||CH

∠MCH=α

∠BCH=β

CH

KB

∠KCH=∠BCH=β

∠MCK=∠MCH-∠KCH=α-β

Now , so . is isosceles, so also. Then . Therefore in the isosceles triangle , the equal angles at and are . On the other hand, from the right angle at , . So , .

α=∠MCH=∠KCH+∠KCM=β+∠KCM

∠KCM=α-β

CKK'

∠K'CM=α-β

∠BCK'=∠BCH+∠HCK+∠KCM+∠MCK'=β+β+(α-β)+(α-β)=2α

BCK'

B

K'

∠CBK'=∠BK'C=π/2-α

B

∠CBK'=π/2-β-δ

α=β+δ

δ=α-β