Question

In the given figure (not to scale) ABC is an isosceles triangle in which AB = AC. AEDC is parallelogram. If $\angle CDF={70}^{\circ }$ and $\angle BFE={100}^{\circ }$ then find $\angle FBA$.

Answer 1

Given, ABC is an isosceles triangle in which AB = AC
AEDC is a parallelogram
$\angle CDF={70}^{\circ }$ and $\angle BFE={100}^{\circ }$
Since AEDC is a parallelogram
$\angle ACD+{70}^{\circ }={180}^{\circ }$
$\Rightarrow \angle ACD={110}^{\circ }\to \left(1\right)$
$\Rightarrow \angle ACD+\angle GCB={180}^{\circ }$ ($\because$ linear pair )
$\Rightarrow \angle GCB={180}^{\circ }-{110}^{\circ }={70}^{\circ }\to \left(2\right)$
$\angle GFD+\angle BFE={180}^{\circ }$ (linear pair)
$\Rightarrow \angle GFD={180}^{\circ }-{100}^{\circ }={80}^{\circ }\to \left(3\right)$
In $\angle BFD,\angle FBD={180}^{\circ }-({80}^{\circ }+{70}^{\circ })={30}^{\circ }$
Since$AB=AC,\angle ABC=\angle ACB$
$\therefore \angle ABC={70}^{\circ }$
$\Rightarrow \angle ABF=\angle ABC-\angle FBD={70}^{\circ }-{30}^{\circ }={40}^{\circ }$