# how to construct angular bisectors for square, pentagon and hexagon?please give me the steps for construction

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*How do you construct an angle bisector?*

The steps will be the same for any angle.

(1) Pick a vertex on the polygon. Name it `/_ABC`

(2) Construct a circle with the center being the chosen vertex, and whose radius is smaller than the side length of the polygon.Mark the intersections of the circle with the two sides of the polygon it intersects; call the intersection between A and B P, and the intersection between B and C Q.

By construction, BP=BQ.

(3) Raise a perpendicular at P perpendicular to `bar(BA)` .

(4) Raise a perpendicular at Q to `bar(BC)`

(5) Mark the intersection of the perpendiculars as R.

(6) Then `bar(BR)` is the angle bisector of `/_ABC` .

Proof: Once you draw `bar(BR)` you will have two triangles; `Delta BPR, Delta BQR`

BP=BQ by construction (radii of circle)

`/_BPR cong /_BQR` as both are right angles by construction.

BR=BR by reflection.

Then the triangles are congruent by HL.

Then PR=QR as corresponding parts of congruent triangles.

By definition, if a point is equidistant from the sides of an angle, then it lies on the angle bisector of that angle, so R is on the angle bisector.