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 post Share on other sites

0 curr3nt 23 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 post Share on other sites

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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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