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In the given figure, side BC of DeltaABC...

In the given figure, side BC of `DeltaABC` is produced to form ray BD and `CE||BA`. Show that `angleACD=angleA+angleB`. Deduce that `angleA+angleB+angleC=180^(@)`.

Text Solution

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`CE||BAandAC` is the transversal.
`:.angleACE=angleA" "....(i)" "["alternate interio"angles]`.
Again, `CE||BAandBD` is the transversal.
`:.angleECD=angleB" "....(ii)" "["corresponding"angles]`
From (i) and (ii), we get
`angleACE+angleECD=angleA+angleB`
`:.angleACD=angleA+angleB`
`impliesangleA+angleB+angleC=angleACD+angleACB`
= a straight angle =`180^(@)`
Hence, `angleACD=angleA+angleBandangleA+angleB+angleC=180^(@)`.
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