## Golden Ratio in an Irregular Pentagon, Construction II

Let A, B, C be three noncollinear points and K complete a parallelogram ABCK. Extend AK to AD such that

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Copyright © 1996-2018 Alexander Bogomolny## Golden Ratio in an Irregular Pentagon, Construction II

Pentagon ABCDE has the diagonals parallel to the sides. In addition, the points of intersection of the diagonals divide each in the golden ratio.

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AD||BC and CE||AB by construction. Also, since

Further, in ΔABD, KM||AB implying BD/BM = AD/AK = φ. Similarly, in ΔBCE, KL||BC implies *BD/BM = BE/BL* so that LM||DE and

In trapezoid DELM, DE/LM = φ and triangles DEK and LMK are similar so that

No consider the four collinear points A, L, K, and D. We have,

(AL + KL) / DK = φ = DK/KL, |

from which we derive

φ = (DK + KL) / AL = DL/AL. |

But, as we already saw, BE/BL = φ and then also BL/EL = φ. Thus in the quadrilateral ABDE the diagonals BE and AD divide each other in equal ratios:

This tells us that ABDE is a trapezoid with BD||AE. Similarly, CD||BE.

Now, points K and L divide AD in the golden ration. Points K and M divide in the golden ration diagonal CE. It's now not difficult to verify that this is also true of points L, M, N, P and the remaining diagonals.

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Copyright © 1996-2018 Alexander Bogomolny