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Given regular pentagon $ABCDE$, prove that $$\frac{DA}{DK} = \frac{DK}{AK}$$

My attempt: By the Triangle Proportionality Theorem, $$\frac{AK}{KD} = \frac{EK}{KB}$$ I'm not too sure about where to go next. Perhaps $\triangle KED \sim \triangle KDB$ by Angle-Angle Similarity?

Pentagon

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  • $\begingroup$ E(K)B is exactly the same as A(K)D. $\endgroup$
    – Mick
    Commented Jul 16, 2019 at 4:44
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    $\begingroup$ The Golden Ratio is present. $\endgroup$ Commented Jul 16, 2019 at 5:08

2 Answers 2

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Since $$\measuredangle DEK=\measuredangle DKE,$$ we obtain $DK=DE=AB$ and it's enough to prove that $$AB^2=AK\cdot AD,$$ for which it's enough to prove that $AB$ is a tangent to the circumcircle of $\Delta DKB,$ for which it's enough to prove that $$\measuredangle KBA=\measuredangle KDB,$$ which is obvious.

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Since it's a regular pentagon, each of the internal angles at $A, B, C, D, E$ have an angle of $108°$. Using that $DC = CB$, then $\triangle DCB$ is isosceles, with $\angle CDB = \angle CBD = 36°$. In fact, using isosceles triangles and sum of angles in a triangle, you get

$$\angle BDA = \angle ADE = \angle BEA = 36° \tag{1}\label{eq1}$$

and

$$\angle DEK = \angle DKE = \angle DAB = \angle DBA = \angle BKA = 72° \tag{2}\label{eq2}$$

Thus, $\triangle DBA \sim \triangle DKE$, so

$$\frac{DA}{DK} = \frac{AB}{EK} \tag{3}\label{eq3}$$

Since $\triangle DKE \cong \triangle BKA$ (by all angles being equal and $DE = AB$), then $EK = AK$. Also, $AB = DE$, due to it being a regular pentagon, and $DE = DK$ due to $\triangle DKE$ being isosceles, so $AB = DK$. Thus, the RHS of \eqref{eq3} can be rewritten as

$$\frac{DA}{DK} = \frac{DK}{AK} \tag{4}\label{eq4}$$

which is what was requested to be proven.

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