Question

Show that any point on the bisector of an angle is equidistant from the arms of the angle.

Answer 1

Solution-

we need to show $BP=CP$

in right angles $△ABP$

$\sin \alpha =\frac{BP}{AP}$...(i)

in right angled $△ACP$

$\sin \alpha =\frac{CP}{AP}$...(ii)

from (i) & (ii)

$\frac{BP}{AP}=\frac{CP}{AP}$

$BP=CP$

Hence Proved.