If a finite straight line segment is divided into two parts so that the rectangle contained by the whole and first part is equal to the square on the other part, prove that the square described on one of the diagonals of the rectangle contained by the whole and the first part is three times the square on other part.
First of all, I am asking this question because I cannot fully understand the question. They are saying to draw a straight line segment and to divide it into two parts. Suppose the two parts are $AB$ and $BC$, and the length of the line segment is $AC$.
I cannot understand the next part of the question. Are they asking for rectangles with diagonals $AB$ and $AC$, or they are asking for rectangles with sides $AB$ and $AC$, and what do they mean by "equal to the square on the other part" ?
Can anyone just help me understand the question? Thank You.
Let $AB=x$ and $BC=y$. Then rectangle contained by the whole and first part is a rectangle with side lengths $x$ and $x+y$; square on the other part is a square with side length $y$, and square described on one of the diagonals of the rectangle contained by the whole and the first part is a square with side length equal to the diagonal of that rectangle.