Question

In the following figure; $AC=CD$; $\angle BAD={110}^0$ and $\angle ACB={74}^0$. Then, $BC>CD$.
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

$\angle ACD={180}^{\circ }-\angle ACB={180}^{\circ }-{74}^{\circ }={106}^{\circ }In△ACD;\angle A=\angle D\angle CAD+\angle ADC={180}^{\circ }-\angle ACD={180}^{\circ }-{106}^{\circ }={74}^{\circ }\angle CAD=\angle ADC={37}^{\circ }In△ABC;\angle ACB={74}^{\circ }\angle BAC=110-\angle CAD=110-37={73}^{\circ }\angle ABC=180-(\angle BAC+\angle ACB)=180-({73}^{\circ }+{74}^{\circ })={180}^{\circ }-{147}^{\circ }={33}^{\circ }\angle BAC>\angle ABCSoBC>ACAC=CD\left(given\right)SoBC>CD$