# 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