# The Plemelj Construction of a Triangle: 6

The Plemelj Construction of a Triangle: 6

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

AB=c

h

C

C

AB

α-β

A

B

Construction

Step 1: Draw a vertical segment of length . Draw two horizontal rays and from and , respectively. From , draw a ray at angle with respect to the ray . Let be the intersection of the rays and .

DE

h

C

ρ

τ

D

E

D

κ

δ

ρ

C

κ

τ

Step 2: Draw a segment of length perpendicular to the ray .

CF

c/2

κ

Step 3: Draw a circle with center and radius . The ray intersects the circle at and .

σ

F

c/2

DF

σ

G

H

Step 4: On , draw the point so that . On , measure out the point so that .

ρ

A

AD=DG

ρ

B

AB=c

Step 5: The triangle satisfies the stated conditions.

ABC

Verification

Draw the circumcircle of the triangle with center . Let the point be the foot of the altitude from . Then . By construction, the power of the point with respect to is , so by Euclid III.37 [1, pp. 15], is tangent to and as angles with orthogonal legs. So .

σ

2

ABC

S

I

C

∠ICS=α-β

D

σ

2

DC=DADB

2

DC

σ

2

δ=∠ICS

δ=α-β