I am stuck on a question from my history of math textbook. Here is the question: The division of a line segment into two unequal parts so that the whole segment will have the same ratio to its larger part that its larger part has to its smaller part is called the golden section. A classical ruler-and-compass construction for the golden section of a segment AB is as follows. At B erect BC equal and perpendicular to AB. Let M be the midpoint of AB, and with MC as a radius, draw a semicircle cutting AB extended in D and E. Then the segment B E laid off on AB gives P, the golden section. (a) Show that 4DBC is similar to 4CBE, whence DB=BCD BC=BE. (b) Subtract 1 from both sides of the equality in part (a) and substitute equals to conclude that AB=AP DAP=P B. (c) Prove that the value of the common ratio in part (b) is (p5 C 1)=2, which is the golden ratio." [Hint: Replace PB by AB AP to see that AB2 AB AP AP2 D0. Divide this equation by AP2 to get a quadratic equation in the ratio AB=AP.] (d) A golden rectangle is a rectangle whose sides are in the ratio (p5 C 1)=2. (The golden rectangle has dimensions pleasing to the eye and was used for the measurements of the facade of the Parthenon and other Greektemples.) Verify that both the rectangles AEFG and BEFC are golden rectangles