# Proving a proportion involving the diagonals of a regular pentagon

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?

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

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.

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.