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 partsso that the whole segment will have the same ratio toits larger part that its larger part has to its smallerpart is called the golden section. A classicalruler-and-compass construction for the golden sectionof a segment AB is as follows. At B erect BC equaland perpendicular to AB. Let M be the midpoint ofAB, and with MC as a radius, draw a semicirclecutting AB extended in D and E. Then the segmentB E laid off on AB gives P, the golden section.(a) Show that 4DBC is similar to 4CBE, whenceDB=BC D BC=B E.(b) Subtract 1 from both sides of the equality inpart (a) and substitute equals to conclude thatAB=AP D AP=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 P B by AB AP to see thatAB2 AB Ð AP AP2 D 0. Divide thisequation by AP2 to get a quadratic equation inthe ratio AB=AP.](d) A golden rectangle is a rectangle whose sidesare in the ratio (p5 C 1)=2. (The goldenrectangle has dimensions pleasing to the eyeand was used for the measurements of thefacade of the Parthenon and other Greektemples.) Verify that both the rectangles AEFG and BEFC are golden rectangles

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Asked May 6, 2019
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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=BC D BC=B E.
(b) Subtract 1 from both sides of the equality in
part (a) and substitute equals to conclude that
AB=AP D AP=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 P B by AB AP to see that
AB2 AB Ð AP AP2 D 0. 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

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Step 1

As per norms , the first three parts a,b and c are answered. The problem pertains to the construction of the point P between given points A and B , which divides AB in the golden ratio.

Step 2

Proof of a)  the triangles are similar as the angles are the same. (recall that DCE is a right angle, as it is the angle in a semicircle, by construction)

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Step 3

b) Using BC=AB, AP=BE, we obtain...

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