# 25. Construct a Triangle Given Its Base, the Difference of the Base Angles and the Length of One of Three Line Segments

25. Construct a Triangle Given Its Base, the Difference of the Base Angles and the Length of One of Three Line Segments

This Demonstration constructs a triangle given the length of its base , the difference between the angles and of the base and the length of one of the following:

ABC

c

AB

δ

α

β

i. the altitude from to the base

C

ii. the bisector from to the base

C

iii. the line segment through the circumcenter from to the base

C

The last two cases are reduced to the first case, since the angle between the bisector and the altitude is , and the angle between the altitude and the line through the center is . So we have a Plemelj triangle, for which we use The Plemelj Construction of a Triangle: 13.

(α-β)/2

α-β

Construction

1. Draw line of length .

AB

c

2. Draw a circle of radius with center , the midpoint of .

σ

1

c/2

M

AB

3. Draw a ray from at an angle to intersect at the point . Let be a point such that the length of is the length of the given altitude and is perpendicular to .

A

δ

σ

1

P

N

MN

MN

AB

4. Draw a circle with center and radius . Let the point be on the circle such that . The point is the intersection of the perpendicular bisector of and the line parallel to through .

σ

2

N

AN

D

AP=AD

C

PD

AB

N

Verification

The angles and are equal to . The angle at is .

∠DAC

∠CAP

β

A

α=β+δ

The first case was posed to Josip Plemelj (1873–1967) by his mathematics teacher in secondary school in 1891 (in Ljubljana, then in the Austro-Hungarian Empire, now in Slovenia).