Question

In the given figure $AE||GF||BD,AB||CG||DF$ and $\angle CHE={120}^{\circ }$, Find $\angle ABC$and $\angle CDE.$
[2 marks]

Answer 1

Since, $BD||AEandCG$ is transversal.
[0.5 marks]

Therefore,$\angle BCH=\angle EHC\Rightarrow \angle BCH={120}^{\circ }$
Again CG ||DF and BD is transversal.
[0.5 marks]

Therefore,$\angle BCH=\angle CDE\Rightarrow \angle CDE={120}^{\circ }$
Also, $AB||CGandBC$ is transversal.
[0.5 marks]

Therefore$\angle ABC+\angle BCH={180}^{\circ }\Rightarrow \angle ABC={180}^{\circ }-{120}^{\circ }\Rightarrow \angle ABC={60}^{\circ }$
[0.5 marks]