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)
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. 