Triangle ABC, with obtuse angle B, is inscribed in a circle. Altitude CH of the triangle is tangent to the circle. If AB=8 and BH=10, find the radius of the circle.
(In reply to
No Subject by jim)
Draw the diameter COC' where O is the center of the circle C' is the point opposite point C.
Draw the tangent at D. Extend HBA to meet this tangent at point H'
Since CC' and HH' are both perpendicular to CH, the quadrilateral CC'H'H is a rectangle. This means that h (the length of HH') is equal to 2r.
By symmetry, AH' = BH =10, so h=28, and r= 14.
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Posted by TomM
on 2005-05-07 20:02:01 |
Solution
