Question

In the following figure, $l_1,l_2,l_3$ are parallel to each other. Find the value of $\angle BDC\text{if}\angle BFE=\angle BCD={50}^{\circ }.$

Answer 1

Given $l_1,l_2,l_3$ are parallel to each other and $\angle BFE=\angle BCD=50°$.

From the figure,
$\angle CAB=90°$
Then, $\angle CAB=\angle FEB=90°$ (Corresponding angles)

Taking the $△BEF$ by angle sum property,
$\angle BEF+\angle BFE+\angle EBF=180°$
$\Rightarrow 90°+50°+\angle EBF=180°$
$\Rightarrow 140°+\angle EBF=180°$
$\Rightarrow \angle EBF=180°-140°=40°$

In triangle ABC, by angle sum property, we have
$\angle ABC+\angle BCA+\angle CAB={180}^{\circ }{40}^{\circ }+\angle BCA+{90}^{\circ }={180}^{\circ }\angle BCA={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Now, $\angle ACD+\angle BDC=180°$ ( Co-interior angles)
$\angle ACD=\angle BCA+\angle BCD={50}^{\circ }+{50}^{\circ }={100}^{\circ }$

$\Rightarrow 100°+\angle BDC=180°$
$\Rightarrow \angle BDC=180°-100°=80°$

Hence, $\angle BDC=80°$.