Given angle #/_ABC# with vertex #B# and two sides #BA# and #BC#. It can be acute or obtuse, or right - makes no difference.

Choose any segment of some length #d# and mark point #M# on side #BA# on a distance #d# from vertex #B#.

Using the same segment of length #d#, mark point #N# on side #BC# on distance #d# from vertex #B#.

Red arc on a picture represents this process, its ends are #M# and #N#.

We can say now that #BM~=BN#.

Choose a radius sufficiently large (greater than half the distance between points #M# and #N#) and draw two circles with centers at points #M# and #N# of this radius. These two circles intersect in two points, #P# and #Q#. See two small arcs intersecting on a picture, their intersection is point #P#.

Chose any of these intersection points, say #P#, and connect it with vertex #B#. This is a bisector of an angle #/_ABC#.

*Proof*

Compare triangles #Delta BMP# and #Delta BNP#.

1. They share side #BP#

2. #BM~=BN# by construction, since we used the same length #d# to mark both points #M# and #N#

3. #MP~=NP# by construction, since we used the same radius of two intersecting circles with centers at points #M# and #N#.

Therefore, triangles #Delta BMP# and #Delta BNP# are congruent by three sides:

#Delta BMP ~=Delta BNP#

As a consequence of congruence of these triangles, corresponding angles have the same measure.

Angles #/_MBP# and #/_NBP# lie across congruent sides #MP# and #NP#.

Therefore, these angles are congruent:

#/_MBP ~= /_NBP#,

that is #BP# is a bisector of angle #/_MBP# (which is the same as angle #/_ABC#).