Prove property: Product of perpendiculars from $P$ to sides of isosceles triagle equals square of perpendicular to base $BC$

Question

$ABC$ is an isosceles triangle with vertex at $A$ and $P$ is any point inside the triangle.If the rectangle contained by perpendicular from $P$ to sides $AB$ and $AC$ is equal to square of the perpendicular from $P$ to base $BC$,then prove that the locus of $P$ is a circle.

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I could not solve this question.I have no idea how to start with.Please help me.Thanks.

Answer 1

Huge courtesy to a friend Appu.S

Interpreting that rectangle means product of perpendicular distances ...

Consider an Isosceles triangle ABC take base along X axis and with its mid point at origin now take the equation of AB as $y=-mx+c$ and AC as $y=mx-c$.

Take a a point $P(x,y)$ now we have

$$y^{2}=\frac{(y+mx-c)(y-mx+c)}{1+m^{2}}$$

Expand and we have a a circle .

To understand the interpretation more clearly take a square which has a side of length $y$ (distance of the point from base) therefore area of square = $y^2$ . Similarly draw a possible case where we can get a rectangle formed and since product of adjacent sides of a rectangle = product of perpendicular distances from our point . We have the area of the rectangle hence formed and we are just equating them as per the question.