[size=150]A right triangle ABC is given with [math]\gamma[/math] = [color=#e06666]90°[/color], along with a point [color=#6aa84f]X[/color] inside the triangle. [br]The feet of the perpendiculars from X divide the sides a, b, c into a[sub]1[/sub], a[sub]2[/sub], and b[sub]1[/sub], b[sub]2[/sub], and c[sub]1[/sub], c[sub]2[/sub]. [br]Squares are constructed on these segments. [br]First, let [color=#6aa84f]X[/color] lie on side c = AB. Vary [color=#6aa84f]X[/color] in such a way that c[sub]1[/sub] and c[sub]2[/sub] remain unchanged [br](e.g., using the up arrow key). [br]Display the auxiliary triangles and use them to justify that a[sub]1[/sub]² + a[sub]2[/sub]² + b[sub]1[/sub]² + b[sub]2[/sub]² ≥ c[sub]1[/sub]² + c[sub]2[/sub]², [br]and that equality holds precisely when X lies on c = AB.[/size]