x̄ - > Sacheri quadrilateral and euclidean proposition

🟡 Geometric Truths in Classical Light

Congruent Diagonals in a Saccheri Quadrilateral

Let us wander through neutral geometry, where postulates pause and theorems bloom from symmetry. Consider a Saccheri quadrilateral \(ABCD\), with:

• Base \( \overline{AB} \)
• Summit \( \overline{CD} \)
• Right angles at \( A \) and \( B \)
• Legs \( \overline{AD} \cong \overline{BC} \)

Draw the diagonals \( \overline{AC} \) and \( \overline{BD} \).

Consider triangles \( \triangle ABC \) and \( \triangle ABD \):
Side: \( \overline{AB} \cong \overline{AB} \) (common side).
Angle: \( \angle ABC \cong \angle BAD \) (both are right).
Side: \( \overline{BC} \cong \overline{AD} \) (given).
Thus, by SAS, \( \triangle ABC \cong \triangle ABD \).

Therefore, \( \overline{AC} \cong \overline{BD} \). The diagonals are equal. A harmony of lines sung across a summit.

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🟢 Question 11: Do Rectangles Exist in Euclidean Geometry?

A rectangle—four corners of perfect rightness. Let us construct one: Start with segment \( \overline{AB} \), erect perpendiculars \( L_1 \) and \( L_2 \) through \( A \) and \( B \).

By the Euclidean Parallel Postulate, these lines are parallel. Pick point \( D \) on \( L_1 \) and construct a line parallel to \( \overline{AB} \), meeting \( L_2 \) at \( C \).

Thus, \( \angle A, B, C, D \) are all right angles.

Therefore, rectangles exist in the Euclidean realm.

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🔵 Question 12: Euclidean Propositions

(a) Supplementary Angles on Parallel Lines

Given: \( \overleftrightarrow{BA} \parallel \overleftrightarrow{CD} \), and \( A \), \( D \) lie on the same side of \( \overleftrightarrow{BC} \).
By Euclid’s 5th Postulate: \( m(\angle ABC) + m(\angle BCD) = 180^\circ \)

(b) Common Perpendicular for Parallel Lines

Given parallel lines \( L_1 \parallel L_2 \), construct perpendicular \( L_3 \) from point on \( L_1 \).
By geometry’s grace, \( L_3 \) intersects \( L_2 \) perpendicularly. A shared spine.

(c) Alternate Interior Angles Are Congruent

Parallel lines \( L_1 \parallel L_2 \), cut by transversal \( T \), form:
\( \angle 1 \cong \angle 2 \), like twin reflections.

(d) Triangle Angle Sum Theorem

In \( \triangle ABC \), draw a line through \( A \) parallel to \( \overline{BC} \):
\( \angle A = \angle 3 \), \( \angle B = \angle 2 \), \( \angle C = \angle 1 \).
Together they lie on a straight line: \[ m(\angle A) + m(\angle B) + m(\angle C) = 180^\circ \]

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🔺 Question 13: The Exterior Angle Theorem with Justification

Any exterior angle of a triangle exceeds either remote interior angle.

Let \( \angle BCD \) be exterior to \( \triangle ABC \).
Construct midpoint \( M \) of \( \overline{BC} \), and reflect point \( A \) across \( M \) to obtain point \( E \) such that \( MA = ME \).

\( \triangle AMB \cong \triangle EMC \) by SAS:
\( \overline{BM} \cong \overline{MC} \), \( \angle AMB \cong \angle EMC \), \( \overline{MA} \cong \overline{ME} \).
So \( \angle ABC \cong \angle ECB \), and since \( \angle BCD = \angle ECB + \angle ECD \), we have: \[ m(\angle BCD) > m(\angle ABC) \]

Q.E.D. The exterior angle casts a longer shadow than those within.

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“Geometry is the art of reasoning well from poorly drawn figures.” — Anonymous