24c. Construct a Triangle Given the Difference of Base Angles, Length of the Altitude from the Base and the Sum of the Lengths of the Other Two Sides
24c. Construct a Triangle Given the Difference of Base Angles, Length of the Altitude from the Base and the Sum of the Lengths of the Other Two Sides
This Demonstration constructs a triangle given the difference of the base angles and , the length of the altitude to the base and the sum of the length of the other two sides and .
ABC
δ
α
β
h
C
s
a
b
\bThis problem is solved by constructing a triangle given the length of a side opposite , the length =sin(δ/2) of an angle bisector at and the angle at .
NBP
s
P
b
P
h
C
P
δ
P
Construction
1. Draw a segment .
NB=s
2. Draw a circle with center and central angle over the chord . Let be the diameter of perpendicular to . Let be the midpoint of .
σ
1
S
2δ
NB
ED
σ
1
NB
M
NB
3. Let be on the extension of so that . Draw the circle with center and radius /2.
F
EN
FN=/2
b
P
σ
2
F
b
P
4. Let intersect at and .
DF
σ
2
G
H
5. Draw a circle with center and radius . Let be the intersection of and . Draw the triangle .
σ
3
D
DH
P
σ
1
σ
3
PNB
6. Let be the intersection of and . Let be the circumcircle of triangle with center .
C
NB
PD
σ
4
NCP
Q
7. Let be the reflection of across .
A
N
PC
Then the triangle satisfies the conditions.
ABC
Verification
Theorem: the segment is the angle bisector of .
CP
∠PNB
In the isosceles triangle , .
∠NCQ=δ/2
∠QNC=π/2-δ/2
Since , . So is on . Therefore, is a tangent of , and the power of with respect to the circle is .
∠NPD=δ/2
∠QNC=π/2
Q
NE
ND
σ
4
D
DPDC=
2
ND
The power of with respect to is . By construction, . So , .
D
σ
2
DHDG=
2
ND
DH=DP
DG=CD
CP=HG=
b
P
Let be the altitude from at . Then . By construction, . So .
CR
C
AB
CR=PCsin(δ/2)
PC=sin(δ/2)
h
C
CR=
h
C
Consider triangle . Its exterior angle at is , and interior angles at other vertices are and . So .
NBP
N
α
β
δ
δ=α-β
Since is symmetric to , . So .
A
N
NC=CA
BC+BC=a+b