Finding a Angle

$\begin{aligned} \frac{1}{CD} - \frac{1}{BD} &= \frac{1}{CD+BD} \\ 1 + \frac{BD}{CD} - \left(1 + \frac{CD}{BD}\right) &= 1 \\ x + -1 + \frac{1}{x} &= 0 && \color{#3D99F6}\text{Setting }x=\frac{BD}{CD} \\ x^2 -x + 1 &= 0 \\ x &= \frac{1+\sqrt{5}}{2} = \phi \end{aligned}$ This is the golden ratio . Thus, it has the property that $\phi = \frac{BC + CD}{CD} = \frac{CD}{BC}$ . This means that our triangle is, in fact, a golden triangle and its apex angle is $\boxed{36^\circ}$ .

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To see why a golden triangle has an apex angle of $36^\circ$ , first consider an isosceles triangle with this apex angle: Bisect one of the base angles and label the resulting segments as shown. By similar triangles, $\frac{a+b}{a} = \frac{a}{b}$ . Thus $\frac{a+b}{a} = \phi$ . Since the leg-base ratio of an isosceles triangle uniquely determines its angles, this is the only triangle with this property.

Let $x=AB=AC=CD$ , $y=BC$ and $θ=∠BAC$ .

Rearranging and simplifying the given equation gives $y^2+xy=x^2$ . ( $BD=x+y$ )

By the cosine rule, $y^2=2x^2(1-cos(θ))$ , which we can reduce to $y^2=4x^2sin^2(θ/2)$ . Thus, $y=2x sin (θ/2)$ .

Substituting this for $y$ in the equation, we form the quadratic equation $4sin^2(θ/2)+2sin(θ/2)-1=0$

$sin(θ/2)= \frac{-1+√5}{4} → θ/2=18$

Therefore, $θ=36$