The diagram shows the circle ABED,with centre O.The tangent at B and D meets AE produced at C.

Prove that DA`xx EC=DC xx DE`

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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)

`rArr DA*EC=DC*DE`

Hence the proof.

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