11.4: Defect

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The defect of triangle $\triangle ABC$ is defined as

$\text{defect} (\triangle ABC) := \pi - |\measuredangle ABC| - |\measuredangle BCA| - |\measuredangle CAB|.$

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.

Exercise $\PageIndex{1}$

Let $\triangle ABC$ be a nondegenerate triangle in the neutral plane. Assume $D$ lies between $A$ and $B$. Show that

$\text{defect} (\triangle ABC) = \text{defect} (\triangle ADC) + \text{defect} (\triangle DBC).$

$截屏2021-02-23 上午10.49.18.png$

Hint: Note that $|\measuredangle ADC| + |\measuredangle CDB| = \pi$. Then apply the definition of the defect.

Exercise $\PageIndex{2}$

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

$\text{defect} (\triangle ABC) = \text{defect} (\triangle AXC).$

$截屏2021-02-23 上午10.50.45.png$

Hint: Show that $\triangle AMX \cong \triangle BMC$. Apply Exercise $\PageIndex{1}$ to $\triangle ABC$ and $\triangle AXC$.

Exercise $\PageIndex{3}$

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

$截屏2021-02-23 上午10.55.37.png$

Show that $B$ and $D$ lie on the opposite sides of $(AC)$. Conclude that

$\text{defect} (\triangle ABC) + \text{defect} (\triangle CDA) = 0.$

Apply Theorem $\PageIndex{1}$ to show that

$\text{defect} (\triangle ABC) = \text{defect} (\triangle CDA = 0$

Use it to show that $\meauredangle CAB = \measuredangle ACD$ and $\measuredangle ACB = \measuredangle CAD$. By ASA, $\triangle ABC \cong \triangle CDA$, and, in particular, $AB =CD$.

(Alternatively, you may apply Exercise 11.3.1)

Advanced Exercise $\PageIndex{4}$

Show that if a neutral plane has a rectangle, then all its triangles have zero defect.

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Exercise \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)

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