Question

In the figure $AB=AC$, $\angle A={48}^o$ and $\angle ACD={18}^o$
Show that $BC=CD$

Answer 1

In $△ABC,AB=AC⟹\angle ACB=\angle ABC=\theta$

In $△ADC,\angle ADC={180}^{\circ }-{48}^{\circ }-{18}^{\circ }={114}^{\circ }$

In $△BCD,\angle BCD=\theta -{18}^{\circ },\angle DCB=\theta ,\angle BDC={180}^{\circ }-{114}^{\circ }={66}^{\circ }$

$\angle BCD+\angle DCB+\angle BDC={180}^{\circ }$

$⟹2\theta ={132}^{\circ }⟹\theta ={66}^{\circ }$

$\angle BDC={66}^{\circ },\angle DCB={66}^{\circ }⟹\angle BDC=\angle DCB⟹BC=CD$