Solution to problem 157. Line AE meets the circumcircle at D. Segment EO meets the circumcircle at F, and the extension of EO meets the circumcircle at G. Taking the circle power of point E with relation to the circumcircle we have EF.EG = ED.EA. But EF.EG = (d-R)(d+R) = d^2 – R^2. By problem 156 we know that ED.EA = 2.R.r1. Hence d^2 – R^2 = 2.R.r1, and d^2 = R^2 + 2.R.r1.
join A to E, A to O
ReplyDeletedraw OG perpendicular to AE
OG² = R² - (AD/2)² ( from tr AOG )
OG² = d² - (AD/2 + DE)² ( from tr OGE )
R² - (AD/2)² = d² - (AD/2)² - AD∙DE - DE²
R² = d² - DE∙( AD + DE )
d² = R² + DE∙AE
from P156 => DE∙AE = 2Rr
d² = R² + 2Rr
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ReplyDeleteSolution to problem 157.
ReplyDeleteLine AE meets the circumcircle at D. Segment EO meets the circumcircle at F, and the extension of EO meets the circumcircle at G.
Taking the circle power of point E with relation to the circumcircle we have EF.EG = ED.EA. But EF.EG = (d-R)(d+R) = d^2 – R^2. By problem 156 we know that ED.EA = 2.R.r1. Hence d^2 – R^2 = 2.R.r1, and d^2 = R^2 + 2.R.r1.