On a circumference, points A and B are on opposite arcs of diameter CD. Line segments CE and DF are perpendicular to AB such that A, E, F and B are collinear in this order. Given that AE=1, find the length of BF.
The answer is that BF=AE
Call the center of the circle O.
If AB is a diameter, then triangles CEO and BFO are congruent right triangles (AAS). OE=OF and subtracting from the radii OA=OB we have BF=AE.
If AB is not a diameter, construct perpendiculars from O to CE, DF, and AB meeting at E', F', and O' respectively. CE'O and BF'O are congruent right triangles (AAS). EE'F'F is a rectangle. Since O bisects E'F', O' bisects EF. Also O' bisects AB. Subtract as before O'A-O'E=O'B-O'F so AE=BF.
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Posted by Jer
on 2025-05-29 08:53:33 |
solution
