WOLFRAM|DEMONSTRATIONS PROJECT

24a. Construct a Triangle Given the Length of the Altitude to the Base, the Difference of Base Angles and the Sum of the Lengths of the Other Sides

​
δ
0.8
s
1.4
length of altitude from C to AB
h
C
0.5
move A'
0.5
show solution
This Demonstration shows a marked-ruler (or verging) construction of a triangle
ABC
given the length
h
C
of the altitude to the base, the difference
δ
of angles at the base and the sum
s
of the lengths of the other two sides.
Construction
From a point
D
draw two rays
ρ
1
and
ρ
2
at an angle
δ
.
Let
H
be on
ρ
1
such that
HD=h
. Let
C
be the intersection of the angle bisector of the angle
ρ
1
ρ
2
and the perpendicular to
ρ
1
at
H
. Thus
∠CDH=δ/2
and
DHC
is a right triangle with hypotenuse
DC
.
Draw the line segment
A'B'
of length
s
with
C
between
A'
and
B'
and
A'
on
ρ
1
.
Move
A'
along
ρ
1
until
B'
is on
ρ
2
; call this point
F
. Then set
A=A'
.
Let
B
be the reflection of
A
in
DC
.
Then
ABC
satisfies the stated conditions.
Proof
In the triangle
DAF
,
∠FDA=δ
,
∠AFD=β
, but the exterior angle
∠FAB=α
, so
δ=α-β
.
Triangles
DCF
and
DCB
are congruent, so
CB=CF
, and
BC+CA=a+b
.