Mathematics 5200/7200

The quadrature of a lune.  Think lunar not loony.  The lune below is the yellow area, the area between the intersection of the two arcs AGD and ADB.

Hippocrates proved that the area of the lune AGDL was equal to the area of the triangle ADC and was therefore quadrable.  He knew that an angle inscribed in a semicircle was a right angle.  He also knew that the square of the diameters of two semicircles were proportional and the Pythagorean theorem.  (By construction DC is perpendicular to AB.  Therefore triangle ADC is congruent to triangle DCB by SAS so AD is equal to DB.)

Proof:

(AB)²= (AD)² + (DB)² = 2(AD)²

Area of semicircle AGD = (AD)² = (AD)²  = 1
Area of semicircle ADB = (AB)² = 2(AC)² = 2

This means that the area of the semicircle AGDE = Area of the quadrant ALDC.

Area of semicircle AGDE - Area of ALDE = Area of quad. ALDC - Area of ALDE
Area of Lune AGDE = Area of triangle ADC.       QED

Hippocrates was also able to prove the quadrature of two other lunes based on the 30, 60 triangle in a similar manner.
 
 

A more modern proof might be:

Let CD=AC=1, therefore AD = square root of 2, and AE = ½ sgrt 2, E is the midpoint of AD.

Area of ALDC = (¶ ·1²)/4  (one quadrant of a circle) = ¶/4

Area of semicircle AGDE = (¶·(sqrt 2/2)²)/2 = ¶/4  (one half of a circle)

Area of triangle ADC = 1/2

The area of the Lune AGDL = Area of the semicircle AGDE - (Area of ALDC - Area of ACD)

                Area of the Lune  = ¼¶ - (¼¶ - ½)
                                                                = ¼¶ - ¼¶ + ½
                                                                =½ = the area of the triangle ADC.     QED

(Dunham)

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