Question

In the given figure, AB = AC, $\angle A={48}^{\circ }$ and $\angle ACD={18}^{\circ }$. Then which of the given options is correct?

• A. BC = BD
• B. BC = CD
• C. BD = DC
• D. BC = 2 BD

Answer 1

The correct option is B BC = CD

In $△ABC$, $\angle BAC+\angle ACB+\angle ABC={180}^{\circ }$
${48}^{\circ }+\angle ACB+\angle ABC={180}^{\circ }$
But $\angle ACB=\angle ABC$ ( $\because$ Given, AB = AC and angles opposite to equal sides of a triangle are equal).
$\therefore 2\angle ABC={180}^{\circ }-{48}^{\circ }$
i.e., $\angle ABC={66}^{\circ }$ $\cdots$ (i)
$\Rightarrow \angle ACB={66}^{\circ }$
$\Rightarrow \angle ACD+\angle DCB={66}^{\circ }$
$\Rightarrow \angle DCB={66}^{\circ }-{18}^{\circ }$ ($\because \angle ACD={18}^{\circ }$)
$\Rightarrow \angle DCB={48}^{\circ }$ $\cdots$ (ii)
Now, in $△DCB$,
$\angle DBC={66}^{\circ }$
(from (i), as $\angle DBC=\angle ABC$)
$\angle DCB={48}^{\circ }$ (from (ii))
$\therefore \angle BDC={180}^{\circ }-{48}^{\circ }-{66}^{\circ }={66}^{\circ }$
$⟹$$\angle BDC=\angle DBC$
Then, BC = CD (sides opposite to equal angles are equal)