Question

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

Answer 1

Since $BCD$ is a straight line,
$\angle ACD={180}^o-{74}^o={106}^o$

From the figure,
$\angle CAD=\angle ADC=x$ (say)

Since sum of angles in a triangle ACD $={180}^o$
${106}^o+x+x={180}^o$
$=>x={37}^o$

Also, $\angle BAC=\angle BAD-\angle CAD={110}^o-{37}^o={73}^o$

Since, $\angle BAC>\angle CAD$, the sides opposite to the angles will also have the same relation.
$=>BC>CD$