WOLFRAM|DEMONSTRATIONS PROJECT

The Plemelj Construction of a Triangle: 3

c

h

C

δ = α-β

0.7

steps

1

2

3

4

additional labels

plot range

1

photograph

This Demonstration constructs a triangle

ABC

given the length

c

of its base

AB

, the length

h

C

of the altitude from

C

to

AB

and the difference

δ=α-β

between the angles at

A

and

B

.

Let

m=

2

c

+

2

(2

h

C

)

.

Construction

Step 1: Draw a right-angled triangle

ABK

with legs of length

AB=c

and

BK=2

h

C

, so that

AK=m

.

Step 2: Construct a circle

σ

with center

A

through

K

(therefore of radius

m

). Construct the ray

ρ

from

B

such that the angle between

ρ

and

BK

is

δ

. Let

K'

be the intersection of

σ

and

ρ

.

Step 3: Let

H

be the midpoint of

BK

, so

BH=

h

C

. Let

M

be the midpoint of

KK'

and let

C

be the intersection of

AM

, the right bisector of

KK'

, with the line through

H

parallel to

AB

.

Step 4: The triangle

ABC

meets the stated conditions.

Verification

Since

AB||CH

,

∠MCH=α

and

∠BCH=β

. The line

CH

is the right bisector of

KB

, so

∠KCH=∠BCH=β

. So

∠MCK=∠MCH-∠KCH=α-β

.

Now

α=∠MCH=∠KCH+∠KCM=β+∠KCM

, so

∠KCM=α-β

.

CKK'

is isosceles, so

∠K'CM=α-β

also. Then

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

. Therefore in the isosceles triangle

BCK'

, the equal angles at

B

and

K'

are

∠CBK'=∠BK'C=π/2-α

. On the other hand, from the right angle at

B

,

∠CBK'=π/2-β-δ

. So

α=β+δ

,

δ=α-β

.