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In the given figure, AB||CD. Prove that ...

In the given figure, `AB||CD`. Prove that `angleBAE-angleECD=angleAEC`

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Through E, draw `EF||AB||CD`.
`EF||CD` and EC is the transversal.
`:. angleCEF+angleECO=180^(@)`.
`AB||EF` and EC is the transversal.
`:. angleCEF+angleECD=180^(@)`.
`AB||EF` and EA is the transversal.
`:. angleBAE+angleAEF=180^(@)`
So, `angleCEF+angleECD=angleBAE+angleAEF`
`implies angleBAE-angleECD=angleCEF-angleAEF=angleAEC`.
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