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A point K is on the perpendicular bisector of a segment with endpoints at H and J. What must be true about point K? It is equidistant from H and J It falls on the h j It is at the midpoint of h j Not enough information is given

Sagot :

Answer:

[tex]K[/tex] is equidistant from [tex]H[/tex] and [tex]J[/tex].

Step-by-step explanation:

Given that the point [tex]K[/tex] which is on the perpendicular bisector of the line segment having endpoints at [tex]H[/tex] and [tex]J[/tex].

The given situation can be represented as the diagram as attached in the answer area.

Referring to the [tex]\triangle HOK, \triangle JOK[/tex]:

[tex]\angle HOK = \angle JOK=90^\circ[/tex] (As it is the perpendicular bisector)

[tex]OH = OJ[/tex] (As it is the perpendicular bisector)

Also, the side [tex]OK[/tex] is the common side.

Therefore by [tex]S-A-S[/tex] congruence, [tex]\triangle HKO\cong \triangle JKO[/tex]

As per the properties of congruent triangles:

Side [tex]HK[/tex] = Side [tex]JK[/tex]

[tex]HK[/tex] and [tex]JK[/tex] are nothing but the distance of the point [tex]K[/tex] from the end points [tex]H[/tex] and [tex]J[/tex] which are proved to be equal to each other.

Therefore, we can conclude that:

[tex]K[/tex] is equidistant from [tex]H[/tex] and [tex]J[/tex].

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