Question

In the given figure, $AB=AC,\angle A={48}^{\circ }$ and $\angle ACD={18}^{\circ }$. Then prove that BC = CD. (3 marks)

Answer 1

In $\Delta ABC$,
$\angle BAC+\angle ACB+\angle ABC={180}^{\circ }$
${48}^{\circ }+\angle ACB+\angle ABC={180}^{\circ }$
But $\angle ACB=\angle ABC(\because Given,AB=AC$ and angles opposite to equal sides of a triangle are equal). (1 mark)
$\therefore 2\angle ABC={180}^{\circ }-{48}^{\circ }$
i.e.,$\angle ABC={66}^{\circ }...\left(i\right)$
$\Rightarrow \angle ACB={66}^{\circ }$
$\Rightarrow \angle ACD+\angle DCB={66}^{\circ }$
$\Rightarrow \angle DCB={66}^{\circ }-{18}^{\circ }$
$(\because \angle ACD={18}^{\circ })$
$\Rightarrow \angle DCB={48}^{\circ }...\left(ii\right)$ (1 mark)
Now, in $\Delta DCB$,
$\angle DBC={66}^{\circ }$
$\left(from\right(i),as\angle DBC=\angle ABC)$
$\angle DCB={48}^{\circ }\left(from\right(ii\left)\right)$
$\therefore \angle BDC={180}^{\circ }-{48}^{\circ }-{66}^{\circ }={66}^{\circ }$
$\Rightarrow \angle BDC=\angle DBC$
Then, BC = CD (sides opposite to equal angles are equal) (1 mark)