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