Question

In $\Delta ABC$, side AC is greater than side AB. If the internal bisector of $\angle A$ meets the opposite side at point D, then $\angle ADC$ is greater than $\angle ADB$.
If the above statement is true then mention answer as 1, else mention 0 if false.

Answer 1

Since $AC>AB$, the angles opposite to the sides will also have the same relation.
$=>\angle B>\angle C$
From the figure, we see that $△ABD$ and $△ACD$ have angles $\angle BAD$ and $\angle CAD$ equal as $AD$ is the angle bisector of $\angle A$
We know that sum of angles in a triangle $={180}^o$
So, in the two triangles, $△ABD$ and $△ACD$
if $\angle BAD=\angle CAD$
$\angle B>\angle C$
$=>\angle ADB<\angle ADC$ as the sum of all angles should be $={180}^o$
Hence, $\angle ADC$ is greater than $\angle ADB$