The Book of Lemmas: Proposition 15
Let AB be the diameter of a circle, AC a side of an inscribed regular pentagon, D the middle point of the arc AC. Join CD and produce it to meet BA produced in E; join AC, DB meeting in F, and draw FM perpendicular to AB. Then, EM = ( radius of circle ).
At the beginning of the proof, you might need to note this fact:
The inscribed angle theorem states that an angle q inscribed in a circle is half of the central angle 2q, that subtends the same arc on the circle.
A consequence of this is: For four consecutive points A,B, C, and D on a circle, ∠ABC + ∠ADC = π.
Let O be the center of the circle and join DA, DM, DO, CB.
Now: ∠ABC = π5 ,
and, ∠ABD = ∠DBC = π10 ,
hence, ∠AOD = π5 .
Further, the triangles FCB, FMB, are equal in all respects.
Therefore, in the triangles DCB, DMB, the sides CB, MB, being equal and BD common, while the angles CBD, MBD are equal,
However, ∠BCD + ∠BAD = two rectangles = π = ∠BAD + ∠DAE = ∠BMD + ∠DMA,
so that, ∠DAE = ∠BCD,
and, ∠BAD = ∠AMD.
AD = MD .
Now, in the triangle DMO,
∠MOD = π5,
∠DMO = 3 π5 ,
(because ∠AMD=∠MAD=∠OAD=∠ODA=2 π5).
Therefore, ∠ODM = π5 = ∠MOD ;
hence, OM = MD .
Again: ∠EDA = supplement of ADC = ∠CBA = π5 = ∠ODM.
Therefore, in the triangles EDA, ODM,
∠EDA = ∠ODM ,
∠EAD = ∠OMD ,
and the sides AD, MD are equal.
Hence, the triangles are equal in all respects and,
EA = MO,
Therefore, EM = AO .
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