Show that the locus of a point which moves in such a way that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distance from other sides, is a circle.

Dear Aman ,

I can do this by coordinate geometry, but I don't see a synthetic (euclidean) proof of it. If the triangle is PQR, with PQ = PR, take P, Q, R to be the points (0,b), (–a,0), (a,0) respectively. The equation of the line PQ is and the equation of PR is . If I understand the question correctly, you want to find the locus of a point X = (x,y) which moves so that the square of its distance from QR is equal to the product of its distances from PQ and PR. With the above choices of P, Q and R, the distance from X to PQ is , and the distance from X to PR is . The distance from X to QR is |y|. So the locus of X is given by the equation . This simplifies to , which represents a circle with centre at and radius . In terms of the geometry of the picture, this circle has its centre C on the perpendicular bisector of QR. It passes through Q and R, and is tangential to PQ and PR. 

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