Question

Prove: If [P(H)] holds, then all Saccheri quadrilaterals have accute top angles.

Answer 1

A Saccheri quadrilateral  is a quadrilateral with two equal sides perpendicular to the base.

• when the summit angles are right angles, the existence of this quadrilateral is equivalent to the statement expounded by Euclid's fifth postulate.
• When the summit angles are acute, this quadrilateral leads to hyperbolic geometry, and
• when the summit angles are obtuse, the quadrilateral leads to elliptical or spherical geometry (provided that also some other modifications are made to the postulates.

Let ABCD be a Saccheri quadrilateral having AB as base, CD as summit and CA and DB as the equal sides that are perpendicular to the base. The following properties are valid in any Saccheri quadrilateral in hyperbolic geometry:

• The summit angles (the angles at C and D) are equal and acute.
• The summit is longer than the base.
• Two Saccheri quadrilaterals are congruent if:
  • the base segments and summit angles are congruent
  • the summit segments and summit angles are congruent.
• The line segment joining the midpoint of the base and the midpoint of the summit:
  • Is perpendicular to the base and the summit,
  • is the only line of symmetry of the quadrilateral,
  • is the shortest segment connecting base and summit,
  • is perpendicular to the line joining the midpoints of the sides,
  • divides the Saccheri quadrilateral into two Lambert quadrilaterals.
• The line segment joining the midpoints of the sides is not perpendicular to either side.

Equations

In the hyperbolic plane of constant curvature , the summit of a Saccheri quadrilateral can be calculated from the leg and the base using the formula

Given ABCD in a neutral geometry, ∠ABD < ∠BDC.

Proof In a neutral geometry, the sum of the measures of the acute angles of a right triangle is less than or equal to 90.

Proof Let 4ABD be a right triangle with right angle at A. Let C be the unique point on the same side of ←→ AD as B with ←→ CD ⊥ ←→ AD and AB ' CD. Then ABCD, and so m(∠ABD) + m(∠ADB) ≤ m(∠BDC) + m(∠ADB). Now B ∈ int(∠ADC) (since ABCD is a convex quadrilateral), so m(∠BDC) + m(∠ADB) = m(∠ADC) = 90. Hence m(∠ABD) + m(∠ADB) ≤ 90.