Loci and Concurrency


   
 
Theorem 2
 
Statement
 
The locus of a point equidistant from two intersecting lines is the pair of lines bisecting the angles formed by the given lines.
 
Part I
 
 
Part II
 
Converse
 
Any point on the bisectors of one of the angles formed by two intersecting lines is equidistant form the lines.
 
 
Part I
 
Data:
 
Let AB and CD be two intersecting lines intersecting at O, forming four angles.
 
 
XOX and YOY bisect the four angles.
 
Let P be a point in the plane such that if PM ^ AB and PN ^ CD, then PM=PN.
 
To prove:
 
 
 
 
Construction:
 
 
Join OP.
 
Proof:
 
Suppose P is a point in the interior of so that PM ^ AB and PN ^ CD and PM=PN, then in right triangles OPM and OPN,
 
OP=OP (common side)
 
PM=PN (given)
 
 
 
 
Hence, P lies on (XOX) U (YOY).
 
Similarly, if P lies in the interior of any of the other three angles then P is on (XOX) U (YOY).
 
Also, P=O then O is the point on both AB and CD; it is
 
equidistant from both AB and CD and so O belongs to (XOX) U (YOY).
 
Thus, in all cases P lies on (XOX) U (YOY).
 
Part II
 
Any point on the bisectors of one of the angles formed by two intersecting lines is equidistant form the lines.
 
Data:
 
Two lines AB and CD intersect at O forming four angles
 
 
XOX and YOY are the lines bisecting the four angles. P is a point of (XOX) U (YOY) such that PM ^ AB and PN ^ CD
 
To prove:
 
PM=PN
 
Proof:
 
Suppose P is different from O, then in triangles OPM and OPN,
 
OP=OP (common side)
 
(given)
 
( given)
 
(AAS congruency theorem)
 
Hence, PM=PN ( CPCT)
 
If P coincides with O, then it is equidistant from AB and CD, since they intersect at O
 
P is equidistant from AB and CD.
 
  
     
   
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