We may use this applet to explore and explain the following property.[br][br][b]For a triangle ABC, if two of its perpendicular bisectors intersect at a point P, this point should also lie on the third perpendicular bisector. [/b]
[icon]/images/ggb/toolbar/mode_showcheckbox.png[/icon]You may try the following steps to explore the figure.[br][list][*]Click the check box to show a movable point Q and the segments connecting it with A, B and C. [br][/*][*]What happens when Q lies on the perpendicular bisector of AB?[br][/*][*]What happens when Q lies on the perpendicular bisector of AC?[br][/*][*]What would you expect to see if Q is moved to P? Why?[br][/*][/list]
Go ahead to move Q and check. Write down your observation here.
You may note that BQ=CQ when Q is moved to P. [br]This is the only position where the 3 segments from Q are equal (that means AQ=BQ=CQ).
Whenever BQ=CQ, Q should lie on the perpendicular bisector of BC. [color=#cccccc](If needed, move Q to some possible positions and check.)[/color][br]Do you know why?
BCQ forms an isosceles triangle. In such case, the perpendicular line from Q to BC should make a pair of congruent triangles and therefore bisect the segment BC. [br]Follow this [url=https://ggbm.at/aWHss4rh]link[/url] to further explore. [br]
Explain why you are sure P is one of those points equally far away from B and C (that means BP=CP).
Since P lies on both perpendicular bisectors, we are sure BP=AP and CP=AP.
So, can you conclude that P must lie on the perpendicular bisector of BC?