Construct a perpendicular to a line

On any line construct a perpendicular line at any point P.

Mark of two points B,C on chosen line equidistant from chosen point P using compass. Increase radius and create two circles centred on B and C with equal radii, which intersect at A and B, with 2AB > BC.

Connect AD which is perpendicular to BC.

Proof.

AD intersects BC at P.

Join BA, BD, CA, CD.

  BA = BD and CA = CD     (radius of circle)

  BA = CA and BD = CD     (construct)               

∠BAD = ∠BDA               (isos. tri.)

similalrly

  ∠CAD = ∠CDA

∴ ∠BAC = ∠BDA

∴ ∇BAC = ∇BDA             (2 sides and inc. ang. eq)

∴ ∠ABC = ∠DBC

    BP = BP

∴ ∇ABP = ∇DBP             (2 sides and inc. ang. eq)

∴ ∠APB = ∠DPB

       = 1 rt. ang.   (ang.s str. line add to 2 rt ang.)

∴ AD is perpendicular to BC

NOTE:

The above was to draw a perpendicular line to another line at a specific point P. However if BC are the end points of a line (segment) then the above constructs the perpendicular bisector of BC since

  ∠APB = 1 rt. ang.

∴ ∠DPC = 1 rt. ang.        (ang.s str. line add to 2 rt ang.)

∴ ∠BAD = 1 rt. ang. - ∠ABC (tri. ang.s add to 2 rt. ang.s) 

       = 1 rt. ang. - ∠ACD (∇ABC isos. AB=AC)

       = ∠CAD

    AB = AC                (construct)

    AP = AP

∴ ∇ABP = ∇ACP              (2 sides & inc. ang. equal) 

∴   BP = CP       

AD is perpendicular bisector of AD .