The golden ratio is defined as:[br][br][math]\frac{a+b}{a}=\frac{a}{b}=\varphi[/math], [math]a>b>0[/math][br][br]Let [math]b=1[/math]. Rearranging term and applying the quadratic formula we can solve for [math]\varphi[/math] directly:[br][br][math]\varphi+1=\varphi^2\Longrightarrow\varphi^2-\varphi-1[/math][br][math]\varphi=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt{5}}{2}[/math][br][br]Note that because we consider [math]\varphi[/math] to be a ratio between strictly positive quantities, we will only take the positive form of this value. Hence [math]\varphi=\frac{1+\sqrt{5}}{2}[/math].

Consider a regular pentagon of side length 1, [math]ABCDE[/math]. Draw lines connecting each interior vertex to form a pentagram in the center of the shape. Observe the smaller pentagon contained within [math]ABCDE[/math], denoted [math]FGHIJ[/math]. [br][br]Let the distance [math]DH=BG[/math] be [math]a[/math]. Constructing a circle of radius [math]a[/math] centered at [math]b[/math] shows that [math]AB=a[/math].[br]Let the distance of [math]HJ[/math] be [math]b[/math]. Note that by constructing a circle of radius [math]b[/math] centered at [math]H[/math], we can see that [math]HB[/math] is also equal to [math]b[/math].[br][br]Consider the two triangles [math]DJH[/math] and [math]DAB[/math]. The triangle proportionality theorem states that if a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. Given that [math]JH\parallel AB[/math], we know that [math]DJH[/math] is proportional to [math]DAB[/math]. Hence, [br][br][math]\frac{DH}{JH}=\frac{DB}{AB}=\frac{a}{b}=\frac{a+b}{a}[/math][br][br]What this informs us of is that the ratio of a diagonal of a regular triangle, [math]EC[/math], and a side length, [math]AB[/math], is the golden ratio.[br][br][math]\frac{EC}{AB}=\varphi[/math]

[math]\varphi=\frac{1+\sqrt{5}}{2}=\frac{1}{2}+\frac{\sqrt{5}}{2}=\frac{1}{2}+\sqrt{\frac{5}{4}}=\frac{1}{2}+\sqrt{\frac{1}{4}+1}=\frac{1}{2}+\sqrt{\left(\frac{1}{2}\right)^2+1^2}[/math][br]Given this representation we can think of the golden ratio as [math]\frac{1}{2}[/math] plus the hypotenuse of a right triangle with side lengths [math]\frac{1}{2}[/math] and [math]1[/math].[br]Begin the construction by drawing a line segment. For the purposes of our construction, we will consider this length to be the unit length of [math]1[/math]. Take this line segment and construct a square. Find the midpoint of the base of the square, and draw a segment from the midpoint to the upper right corner of the square. This is the hypotenuse of a [math]\frac{1}{2}[/math] by [math]1[/math] triangle and is thus equal to [math]\frac{\sqrt{5}}{2}[/math].[br]Construct a circle centered at the midpoint, of radius equal to the constructed line segment. Extend the original base of the square to intersect with the circle. This distance will be of [math]\frac{1+\sqrt{5}}{2}[/math].

Take the above construction of the golden ratio, and construct two circles of radius [math]a[/math] centered at [math]D[/math] and [math]H[/math]. Their upper intersection will form the top of the pentagon.

Draw two new circles of radius [math]a+b[/math] centered at [math]D[/math] and [math]H[/math]. Their intersections with the original circles represent the bottom diagonal lines for the pentagon, and form the last two vertices of the pentagon.

Thus a regular pentagon constructed with only Euclidean construction.