Question

Construct the locus of point equidistant from $BA$ and $BC$. Hence construct a circle touching the three sides of the triangle internally.

Answer 1

1)Drawn a straight line $AB=7$ cm with the help of ruler.
2)With the help of compass drawn an arc from $A$ and at the point where it cuts $AB$ from that point made another arc drawn an arc cutting the previous arc.
3)From $A$ drawn a straight line joining the arc and extend it to $M$.
4)With the help of ruler measured $5$ cm and mark it as $AC$.
5)Joined $BC$ and we get the required triangle.
6)From $A$ drawn an arc and make it cut on $AC$ and $AB$ and from the point it cuts $AC$ and $AB$ drawn arc cutting each other and extend a line from point $A$ extend a line to the point point of intersection of two arc
7)Similarly we do for $B$ and the point where the two line intersect denoted as $O$.
8)Made a perpendicular from $O$ on $AB$ this perpendicular will be radius and taking $O$ as centre we draw a circle this is our incircle.
9)And $BN$ is our locus of points equidistant from two lines BA and BC.
We need to construct a circle inscribed in triangle that is incircle it can be done by making angle bisector of two sides the point where it intersect will be incentre. The centre of required circle.
The angle bisector is the locus where points are equidistant from two sides.