Question

In the given figure, $AE\parallel BD$, $AC\parallel ED$ and $AB=AC$. Find $\angle a$.

• A. ${63}^o$
• B. ${64}^o$
• C. ${62}^o$
• D. None of the above

Answer 1

Answer: (B)

${64}^o$

Here, $\angle EDC={58}^{\circ }$ (Vertically opposite angles)
Given $AE\parallel CD$ so,

$⟹$ $\angle c=\angle EDC={58}^{\circ }$ [Corresponding angles]
Also, $AC\parallel ED$
$⟹$ $\angle ACB=\angle EDC={58}^{\circ }$ [Corresponding angles)
Since, $AB=AC$
$⟹$ $\angle ABC=\angle ACB={58}^o$ [Isosceles triangle property]

Now, in $△ABC$,
$\angle ABC+\angle ACB+\angle BAC={180}^o$
$⟹$ ${58}^o+{58}^o+\angle a={180}^o$

$⟹$${116}^o+\angle a={180}^o$
$⟹$ $\angle a={180}^o-{116}^o$
$⟹$ $\angle a={64}^o$

Hence, option $B$ is correct.