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The diagram shows the circle ABED,with centre O.The tangent at B and D meets AE...
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Consider triangles `DeltaADC` and `DeltaDEC` ,
<DCE=<DCA (common angle),
<EDC=`90^o` -<ODE (because CD is a tangent, OD radius)
=<ADO (<ADE is a semicircular angle, hence right angle; and <ADE=<ADO+<ODE)
=<DAO (For the triangle DeltaDAO, DO=OA=radius of the circle, hence it is isosceles, so, <DAO=<ADO)
Two angles being equal, the two triangles `DeltaADC` and `DeltaDEC` are equialgular, hence similar.
Therefore, their corresponding sides are in the same ratio.
i.e. `(DE)/(DA)` (sides opposite angle <C)`=(EC)/(DC)` (sides opposite to equal angles <EDC and <DAC)
Hence the proof.
Posted by llltkl on August 31, 2013 at 3:12 PM (Answer #1)
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