Question

In the figure AB = AC, BC = CD and DE || BC. Find the measure of $\angle x$.

• A. ${54}^o$
• B. ${61}^o$
• C. ${58}^o$
• D. ${72}^o$

Answer 1

Answer: (C)

${58}^o$

Given : $AB=AC$

$\therefore \angle ABC=\angle ACB$

$\angle ABC+\angle ACB={180}^o-\angle BAC={122}^o$

$\therefore \angle ABC=\angle ACB=\frac{122}{2}={61}^o$

In $\Delta BCD,$

$BC=CD$

$\therefore \angle DBC=\angle CDB={61}^o$ ..... [Angles opposite to equal sides are equal]

$DE\parallel BC$

$\angle EDB+\angle DBC={180}^o$

$\therefore x+61+61=180$

$x=180-122={58}^o$