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In the given figure, `BA||ED` and `BC||EF`. Show that `angleABC=angleDEF`.

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In the given figure, `BA||ED` and `BC||EF`. Show that `angleABC=angleDEF`.
image
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Product DE to meet BC at P.
`BA||PD` and BPC is the transcersal.
`:. angleABC=angleDPC` [corresponding `angles`]
`EF||BC` and `DP` is the transversal.
`:. angleDPC=angleDEF` [corresponding `angles`].
Hence, `angleABC=angleDEF`.
image
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