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In the figure, chord AB || tangent DE. ...

In the figure, chord `AB ||` tangent DE. Tangent DE touches the circle at point C then prove AC = BC.

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Seg `AB || `line DE and line AC is the transversal,
`:. /_BAC ~= /_ACD ` …(Alternate angles ) …(1)
`/_ACD ~= /_ABC ` ….(Tangent decant theorem ) ….(2)
`:.` from (1) and (2) , we get ,
`/_ BAC ~= /_ABC `
`:. ` seg `AC ~= /_ABC `
`:.` seg `AC ~= ` seg BC ....(Converse of isosceles triangle theorem )
`:. AC = BC `
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