Question

In the following figure:
$AC=CD;\angle BAD={110}^o$ and $\angle ACB={74}^o$.
Prove that: $BC>CD$.

Answer 1

$\angle ABC+73+74=180$

$\angle ABC=180-147$

$\angle ABC={33}^o$

$\angle ACD=180-74=106$

$2x+106=180$

$2x=180-106=74$

$x=37$

$\angle CAB=110-37={73}^o$

According to their angle we can prove $BC>CD$.