In the given figure, `AB||CD, angleGED=125^(@)` and `EF bot CD`. Find `angleAGE, angleGEF` and `angleFGE`.

Explanation

`AB||CD` and GE is the transversal.
`:. angleAGE=angleGED=125^(@)`
[alternate interior `angles`]
and `angleGEF=(angleGED-angleFED)=(125^(@)-90^(@))=35^(@)`.
Now, `AB||CD` and EF is the transversal.
`:. angleBFE+angleFED=180^(@)` [sum of int. `angles`]
`implies angleBFE+90^(@)=180^(@) implies angleBFE=90^(@)`.
Now, AB is a straight line and EF stands on it.
`:. angleGFE+angleBFE180^(@) implies angleGFE+90^(@)=180^(@)`
`implies angleGFE=(180^(@)-90^(@))=90^(@)`.
We kanow that the sum of the angles of a triangle is `180^(@)`.
In `Delta GEF`, we have
`angleGEF+angleEFG+angleFGE=180^(@)`
`implies 35^(@)+90^(@)+angleFGE=180^(@)`
`implies angleFGE=(180^(@)-125^(@))=55^(@)`
Hence, `angleAGE=125^(@), angleGEF=35^(@)` and `angleFGE=55^(@)`.