Theorem 2


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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,

 
Suppose P is a point in the interior of so that PM

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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