In the given figure, if AB || CD,EF || BC, =65^ and =35^ , find .

Question

In the given figure, if $\text{AB }||\text{ CD},\text{EF }||\text{ BC},\angle\text{BAC}=65^\circ$ and $\angle\text{DHF}=35^\circ,$ find $\angle\text{AGH}.$

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Answer

In the given figure, if $\text{AB }||\text{ CD},\text{EF }||\text{ BC},\angle\text{BAC}=65^\circ$ and $\angle\text{DHF}=35^\circ$ We need to find $\angle\text{AGH}.$

Here, GF and CD are straight lines intersecting at point H, so using the property, "vertically opposite angles are equal", we get,$\angle\text{DHF}=\angle\text{GHC}$
$\angle\text{GHC}=35^\circ$
Further, as AB || CD and AC is the transversal Using the property, "alternate interior angles are equal"$\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{BAC}=65^\circ$
Further applying angle sum property of the triangle In $\triangle\text{GHC}$$\angle\text{DHF}+\angle\text{HCG}+\angle\text{CGH}=180^\circ$
$\angle\text{CGH}+35^\circ+65^\circ=180^\circ$
$100^\circ+\angle\text{CGH}=180^\circ$
$\angle\text{CGH}=180^\circ-100^\circ$
$\angle\text{CGH}=80^\circ$
Hence, applying the property, "angles forming a linear pair are supplementary" As AGC is a straight line$\angle\text{CGH}+\angle\text{AGH}=180^\circ$
$\angle\text{AGH}+80^\circ=180^\circ$
$\angle\text{AGH}=180^\circ-80^\circ$
$\angle\text{AGH}=100^\circ$
Therefore,$\angle\text{AGH}=100^\circ$