MHB What is the Multiplication Rule for AD and CD in a Circle with a Diameter BE?

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The discussion explains the multiplication rule for segments AD and CD in a circle with diameter BE. It highlights that the circle centered at point P, with radius PA equal to PB, passes through point C, where the angle at the center is twice that at the circumference. By extending BP to form the diameter BE, the relationship AD·CD = BD·DE is established. The specific calculation shows that AD·CD equals 7, derived from the values 1 and (8-1). This illustrates the geometric principles involved in the multiplication rule within the context of circle geometry.
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[sp]The circle centred at $P$ with radius $PA = PB$ passes through $C$ (angle at centre = twice angle at circumference). Extend $BP$ to form a diameter $BE$ of the circle. Then $AD\cdot CD = BD\cdot DE = 1\cdot (8-1) = 7$.[/sp]
 
Opalg said:
[sp]The circle centred at $P$ with radius $PA = PB$ passes through $C$ (angle at centre = twice angle at circumference). Extend $BP$ to form a diameter $BE$ of the circle. Then $AD\cdot CD = BD\cdot DE = 1\cdot (8-1) = 7$.[/sp]
very good !
 

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