The Plemelj Construction of a Triangle: 10

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

between the angles

α

at

A

and

β

at

B

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

Let

m=

2

c

+

2

(2

h

C

)

.

Construction

Draw a right-angled triangle

ABK

with legs of length

AB=c

and

AK=2

h

C

, so that

BK=m

.

Step 1: Construct a circle

σ

with center

B

through

K

and, therefore, of radius

m

. Construct the ray

ρ

from

A

such that the angle between

ρ

and

AK

is

δ

. Let

K'

be the intersection of

σ

and

ρ

.

Step 2: Let

H

be the midpoint of

AK

, so that

AH=

h

C

. Let

M

be the midpoint of

KK'

. Let

C

be the intersection of the perpendicular bisector of

KK'

and the line through

H

parallel to

AB

.

Step 3: The triangle

ABC

meets the stated conditions.

Verification

Since

AB||HC

and

MB

is the perpendicular bisector of

KK'

,

∠MCH=β

and

∠ACH=α

. So

∠MCK=∠KCH-∠HCM=α-β

. The triangle

CKK'

is isosceles, so

∠K'CM=α-β

also and

∠K'CK=2(α-β)

, and

∠ACK'=2α-2(α-β)=2β

.

Since

K'AC

is isosceles,

∠ACK'=π-2(π/2-α+δ)=2(α-δ)

.

So

2β=2(α-δ)

and

δ=α-β

.