For a circle with center [i]A[/i] through [i]B[/i] and a free point [i]P[/i], we consider the point [i]D[/i] intersection of the circle and the line [i]PA[/i]. When are the lines [i]AD[/i] and [i]BD[/i] perpendicular?

We obtain the answer by typing [code]LocusEquation[ArePerpendicular[a,d],P][/code] (here the result can be displayed by clicking on the right check box). We get the line [i]AB[/i] passing through [i]A[/i] and [i]B[/i] includes three interesting types of different situations:[br][list][*]For [i]P[/i] in [i]AB[/i] different from [i]A[/i] and [i]B[/i] we obtain the degenerate case in which the intersection point [i]D[/i] coincides with [i]B[/i] and hence the property is trivially true.[br][/*][*]Something similar happens if [i]P[/i] is equal to [i]A[/i]. This is again a degenerate case since the line [i]PA[/i] does not really exist, and hence the point [i]D[/i] is undefined.[br][/*][*]Something very different, however, happens when [i]P[/i] is equal to [i]B[/i]. In this case [i]D[/i] is the "opposite" point to [i]B[/i] along the diameter [i]AB[/i], and hence line [i]AD[/i] is not perpendicular to line [i]BD[/i] but equal and hence the property is false. It is important to emphasize that this condition [i]P[/i]=[i]B[/i] is a particular case of the locus set of necessary conditions that is not sufficient.[/*][/list]