27b. Construct a Triangle Given Its Perimeter, an Angle and the Length of the Altitude to the Side Opposite the Angle
27b. Construct a Triangle Given Its Perimeter, an Angle and the Length of the Altitude to the Side Opposite the Angle
This Demonstration constructs a triangle given the perimeter , the angle at and the length of the altitude from to the base .
ABC
p
γ
C
h
C
C
AB
This construction uses a previous method to construct a triangle given the length of the base , the opposite angle and the length of the altitude to the base.
CDE
ED
C
h
C
Construction
1. Draw a triangle given the length of the base , the angle opposite the base and the length of the altitude from .
CDE
p
ED
γ/2+π/2
h
C
C
2. Let be the intersection of and the perpendicular bisector of . Similarly, let be the intersection of and the perpendicular bisector of .
A
DE
EC
B
DE
CD
3. Then the triangle satisfies the conditions.
ABC
Verification
By step 1, . Also, since is isosceles, . Similarly, . So .
∠ECD=γ/2+π/2
EAC
∠ECA=α/2
∠BCD=β/2
∠ACB=∠ECD-∠ECA-∠BCD=γ/2+π/2-α/2-β/2
Since , .
α+β+∠ACB=π
1/2∠ACB=π/2-(α/2+β/2)
Subtracting, .
∠ACB-1/2∠ACB=[γ/2+π/2-α/2-β/2]-[π/2-(α/2+β/2)]
Simplifying, .
∠ACB=γ
By step 1 and isosceles triangles, is the perimeter of .
ED=p
ABC