Question 52048
How do i prove the characterization of the internal angle bisector theorem: Let (P) be in the interior of an angle with arms L1 and L2 then (P) is on the angle bisector if and only if d(P,L1) = d(P,L2)
YOUR WRITE UP IS NOT CLEAR..
LET AOB BE THE ANGLE....AO IS L1 AND BO IS L2
LET OX BE THE INTERNAL BISECTOR OF ANGLE AOB...
P IS A POINT ON OX IF AND ONLY IF PQ = PR WHERE PQ AND PR ARE PERPENDICULAR DISTANCES OF P FROM L1 AND L2...IF SO THE PROOF IS AS FOLLOWS...
CASE 1 
P IS ON OX 
TPT...PQ=PR
IN TRIANGLES PQO AND PRO
PO=PO
ANGLE POQ=ANGLE POR..PO IS ANGLE BISECTOR
ANGLE PQO =90 = ANGLE PRO ..PQ,PR ARE PERPENDICULAR DISTANCES 
HENCE THE 2 TRIANGLES ARE CONGRUENT.
HENCE PQ=PR.PROVED
CASE 2...
PQ=PR...
TPT OP IS ANGLE BISECTOR..

IN TRIANGLES PQO AND PRO
PO=PO
PQ=PR.GIVEN
ANGLE PQO =90 = ANGLE PRO ..PQ,PR ARE PERPENDICULAR DISTANCES 
HENCE THE 2 TRIANGLES ARE CONGRUENT.
HENCE ANGLE POQ=ANGLE POR...PROVED