11.4: Defect
( \newcommand{\kernel}{\mathrm{null}\,}\)
The defect of triangle △ABC is defined as
defect(△ABC):=π−|∡ABC|−|∡BCA|−|∡CAB|.
Note that Theorem 11.3.1 states that the defect of any triangle in a neutral plane has to be nonnegative. According to Theorem 7.4.1, any triangle in the Euclidean plane has zero defect.
Let △ABC be a nondegenerate triangle in the neutral plane. Assume D lies between A and B. Show that
defect(△ABC)=defect(△ADC)+defect(△DBC).
- Hint
-
Note that |∡ADC|+|∡CDB|=π. Then apply the definition of the defect.
Let ABC be a nondegenerate triangle in the neutral plane. Suppose X is the reflection of C across the midpoint M of [AB]. Show that
defect(△ABC)=defect(△AXC).
- Hint
-
Show that △AMX≅△BMC. Apply Exercise 11.4.1 to △ABC and △AXC.
Suppose that ABCD is a rectangle in a neutral plane; that is, ABCD is a quadrangle with all right angles. Show that AB=CD.
- Hint
-
Show that B and D lie on the opposite sides of (AC). Conclude that
defect(△ABC)+defect(△CDA)=0.
Apply Theorem 11.4.1 to show that
defect(△ABC)=defect(△CDA=0
Use it to show that \meauredangleCAB=∡ACD and ∡ACB=∡CAD. By ASA, △ABC≅△CDA, and, in particular, AB=CD.
(Alternatively, you may apply Exercise 11.3.1)
Show that if a neutral plane has a rectangle, then all its triangles have zero defect.