Guest Posted March 1, 2011 Report Share Posted March 1, 2011 start with a 3 4 5 right triangle; ABC. construct a circle with the diameter along the 3 side. (AB) it will intersect the side of length 5 at point D. according the the secant tangent property, we have BC^2 = AD*DC now construct anther circle with diameter along the 4 side. (BC) this circle will also intersect the side of length 5 at D. therefore by the same property, we have AB^2 = AD*DC. therefore AB^2 = BC^2, or AB = BC or 3 = 4. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted March 1, 2011 Report Share Posted March 1, 2011 (edited) Isn't the theorem AB^2 = AD * AC => 3^2 = 1.8 * 5 BC^2 = CD * AC => 4^2 = 3.2 * 5 edit - fixed spelling Edited March 1, 2011 by curr3nt Quote Link to comment Share on other sites More sharing options...
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Guest
start with a 3 4 5 right triangle; ABC.
construct a circle with the diameter along the 3 side. (AB)
it will intersect the side of length 5 at point D.
according the the secant tangent property, we have BC^2 = AD*DC
now construct anther circle with diameter along the 4 side. (BC)
this circle will also intersect the side of length 5 at D.
therefore by the same property, we have AB^2 = AD*DC.
therefore AB^2 = BC^2, or AB = BC or 3 = 4.
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