In the diagram, triangle ABC is inscribed in the circle and FEG is a tangent to the circle at E such that EF=EC.FBC is a straight line and EA intersects BC at D.

Show that angle CAD= 2 angle BAD,thus show that `FB-:FE = CE-:CF.`

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Since FE=CE `DeltaEFC` is isosceles with vertes angle E; thus `m/_ECF=m/_EFC` .

The measure of an inscribed angle is 1/2 the intercepted arc so `m/_C=1/2"arc"BE`

The measure of an angle formed by a secant and a tangent is 1/2 the difference of the intercepted arcs -- so `m/_F=1/2("arc"CE-"arc"BE")`

Then `1/2"arc"BE=1/2("arc"CE-"arc"BE)==>"arc"CE=2"arc"BE`

Again the measure of an inscribed angle is 1/2 the intercepted arc. Thus `m/_BAE=1/2"arc"BE` and

`m/_CAE=1/2"arc"CE`

So `m/_CAD=2m/_BAD` .

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There are a number of approached from here.

Let `m/_BAD=x,m/_CAD=2x` . Then `m/_BFE=x` and `m/_BCE=x` ``, and `m/_BEF=x` since they all intercept the same arc.

Then `DeltaFBE` ~ `DeltaFEC` by AA~.

The sides of similar triangle are in proportion so

`"FB"/"CE"="FE"/"CF"` Exchanging the means of the proportion yields:

`"FB"/"FE"="CE"/"CF"` as required.

** Using the fact that CE=FE we can write as `FB*CF=(FE)^2`

which says that the length of a tangent segment is the geometric mean between the external secant segment and the secant segment drawn from the same point. **

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