# 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 of an angle bisector at and the angle at .

NBP

s

P

b=h/sin(δ/2)

P

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 .

F

EN

FN=b/2

P

σ

2

F

b/2

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=ND

2

The power of with respect to is . By construction, . So , .

D

σ

2

DHDG=ND

2

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=h/sin(δ/2)

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