Taking a scale of $2 \; cm$ as $2$ units on both the axes, on a graph paper plot the points $A \left(1, 2\right)$, $B \left(8, 3\right)$, $C \left(7, 8\right)$ and $D \left(3, 7\right)$.
Complete the quadrilateral $ABCD$.
Find the point $P$ inside the quadrilateral such that $AP = BP$ and $P$ is also equidistant from $AB$ and $AD$.
Record the length $PB$.
The points $A \left(1, 2\right)$, $B \left(8, 3\right)$, $C \left(7, 8\right)$ and $D \left(3, 7\right)$ are plotted on a graph paper and the quadrilateral $ABCD$ is drawn.
Given: $\;$ $AP = BP$
$\implies$ $P$ lies on the perpendicular bisector of the line $AB$.
[The locus of a point, which is equidistant from two fixed points, is the perpendicular bisector of the line segment joining the two fixed points.]
$EF$ is the perpendicular bisector of the line segment $AB$.
Given: $\;$ $P$ is equidistant from $AB$ and $AD$.
$\implies$ $P$ lies on the angle bisector of lines segments $AB$ and $AD$.
[The locus of a point equidistant from two intersecting lines is the bisector of the angle between the lines.]
$AG$ is the angle bisector of $\angle DAB$.
$EF$ and $AG$ intersect at point $P$, which is the the point inside the quadrilateral such that $AP = BP$ and is equidistant from $AB$ and $AD$.
Length of $PB = 6.5 \; cm$