
# 13.4: Hyperbolic triangles


Theorem $$\PageIndex{1}$$

Any nondegenerate hyperbolic triangle has a positive defect.

Proof

Fix an h-triangle $$ABC$$. According to Theorem 11.3.1,

$\text{defect}(\triangle_hABC)\ge 0$

It remains to show that in the case of equality, $$\triangle_hABC$$ degenerates.

Without loss of generality, we may assume that $$A$$ is the center of the absolute; in this case $$\measuredangle_h CAB = \measuredangle CAB$$. Yet we may assume that

$$\measuredangle_h CAB$$,    $$\measuredangle_h ABC$$,    $$\measuredangle_h BCA$$, $$\measuredangle ABC$$,    $$\measuredangle BCA \ge 0.$$

Let $$D$$ be an arbitrary point in $$[CB]_h$$ distinct from $$B$$ and $$C$$. From Proposition 9.6.1, we have

$$\measuredangle ABC-\measuredangle_h ABC \equiv \pi-\measuredangle CDB \equiv \measuredangle BCA-\measuredangle_h BCA.$$

From Exercise 7.4.2, we get that

$$\text{defect}(\triangle_hABC)=2\cdot(\pi-\measuredangle CDB).$$

Therefore, if we have equality in 13.4.1, then $$\measuredangle CDB=\pi$$. In particular, the h-segment $$[BC]_h$$ coincides with the Euclidean segment $$[BC]$$. By Exercise 12.1.3, the latter can happen only if the h-line $$(BC)_h$$ passes thru the center of the absolute ($$A$$); that is, if $$\triangle_hABC$$ degenerates.

The following theorem states, in particular, that nondegenerate hyperbolic triangles are congruent if their corresponding angles are equal. In particular, in hyperbolic geometry, similar triangles have to be congruent.

Theorem $$\PageIndex{2}$$ AAA congruence condition

Two nondegenerate h-triangles $$ABC$$ and $$A'B'C'$$ are congruent if $$\measuredangle_hABC = \pm \measuredangle_hA'B'C'$$, $$\measuredangle_hBCA = \pm \measuredangle_h B'C'A'$$ and $$\measuredangle_hCAB = \pm \measuredangle_hC'A'B'$$.

Proof

Note that if $$AB_h=A'B'_h$$, then the theorem follows from ASA.

Assume the contrary. Without loss of generality, we may assume that $$AB_h<A'B'_h$$. Therefore, we can choose the point $$B''\in [A'B']_h$$ such that $$A'B''_h=AB_h$$.

Choose an h-half-line $$[B''X)$$ so that

$$\measuredangle_h A'B''X=\measuredangle_h A'B'C'.$$

According to Exercise 11.5, $$(B''X)_h\parallel(B'C')_h$$.

By Pasch’s theorem (Theorem 3.12), $$(B''X)_h$$ intersects $$[A'C']_h$$. Suppose that $$C''$$ denotes the point of intersection.

According to ASA, $$\triangle_h ABC\cong\triangle_h A'B''C''$$; in particular,

$\text{defect}(\triangle_h ABC)=\text{defect}(\triangle_h A'B''C'').$

Applying Exercise 11.11 twice, we get that

$\begin{array} {rcl} {\text{defect} (\triangle_h A'B'C')} & = & {\text{defect} (\triangle_h A'B''C'') +} \\ {} & + & {\text{defect}(\triangle_h B''C''C') + \text{defect} (\triangle_h B''C'B').} \end{array}$

By Theorem 13.7, all the defects have to be positive. Therefore

$$\text{defect} (\triangle_h A'B'C') > \text{defect} (\triangle_h ABC).$$

On the other hand,

$$\begin{array} {rcl} {\text{defect} (\triangle_h A'B'C')} & = & {|\measuredangle_h A'B'C'| + |\measuredangle_h B'C'A'| + |\measuredangle_h C'A'B'| =} \\ {} & = & {|\measuredangle_h ABC| + |\measuredangle_BCA| + |\measuredangle_h CAB|} \\ {} & = & {\text{defect} (\triangle_h ABC)} \end{array}$$

Recall that a bijection from a h-plane to itself is called angle preserving if

$$\measuredangle_h ABC= \measuredangle_h A'B'C'$$

for any $$\triangle_h ABC$$ and its image $$\triangle_h A'B'C'$$.

Exercise $$\PageIndex{1}$$

Show that any angle-preserving transformation of the h-plane is a motion.

Hint

Apply AAA-congruence condition (Theorem $$\PageIndex{2}$$)