Question

From the following figure, AB > CD.

If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In triangle, $ABC$,since $AB=AC$, the angles opposite to each other are equal
$=>\angle ABC=\angle ACB={70}^o$
Since
$BCD$ is a straight line,
$\angle ACD={180}^o-{70}^o={110}^o$
Since
sum of angles in a triangle ACD $={180}^o$

$\angle ACD+\angle ADC+\angle CAD={180}^o$
$=>{110}^o+{40}^o+\angle CAD={180}^o$
$=>\angle CAD={30}^o$
Now, $\angle ACB>\angle CAD$

This
means, the sides opposite to the angles will also have the same relation.
$=>AB>CD$