A geometry problem involving geometric mean

Question

$ABCD$ is a quadrilateral inscribed in a circle of center $O$. Let $BD$ bisect $OC$ perpendicularly. $P$ is a point on the diagonal $AC$ such that $PC=OC$. $BP$ cuts $AD$ at $E$ and the circle circumscribing $ABCD$ at $F$.

Prove that $PF$ is the geometric mean of $EF$ and $BF$.

Answer 1

By construction, points $O$, $B$, $D$, $P$ lie on $\bigcirc C$, and $\stackrel{\frown}{BCD} = \stackrel{\frown}{BOD} = 120^\circ$.

Note that $\angle DBF \cong DCF$ (marked "$\theta$") as inscribed angles of $\bigcirc O$ subtending $\stackrel{\frown}{DF}$. Likewise, $\angle ACF\cong \angle ADF$ (marked "$\phi$").

Further, since $\angle BFD$ subtends $\stackrel{\frown}{BCD}$ of $\bigcirc O$, and since $\angle BPD$ subtends major arc $\stackrel{\frown}{BD}$ of $\bigcirc C$, we have

$$\angle BFD = \frac{1}{2}\stackrel{\frown}{BCD} =60^\circ$$ $$\angle DPF = 180^\circ - \angle DPB = 180^\circ - \frac{1}{2}\left(360^\circ-\stackrel{\frown}{BOD}\right) = 60^\circ$$ This implies that $\triangle PDF$ is equilateral. Moreover, $\square PCDF$ is a kite, with diagonal $\overline{CF}$ bisecting $\angle C$, so that $\theta = \phi$.

Consequently, $\triangle FDE \sim \triangle FBD$, whereupon $$\frac{|\overline{EF}|}{|\overline{DF}|} = \frac{|\overline{DF}|}{|\overline{BF}|}\qquad\to\qquad |\overline{DF}|^2 = |\overline{BF}||\overline{EF}|$$

Since $\overline{DF}\cong\overline{PF}$ in equilateral $\triangle PDF$, we have our result. $\square$

Answer 2

This is a coordinate-based solution. Without loss of generality, you can choose your coordinate system such that the circle $\bigcirc ABCD$ becomes the unit circle and $C$ has coordinates $(1,0)$. Then, in order to simplify the formulation, we start by choosing rational coordinates for point $P$:

$$P = (2t^2,2t)/(t^2+1)$$

The coordinates are chosen in such a way that they describe any point on the circle $(x-1)^2+y^2=1$ except for the point $(2,0)$ (which corresponds to $t\to\infty$). For any point $P$ on that circle you can compute $t=y/(2-x)$.

From these coordinates, everything else can be computed:

\begin{align*} O&=(0,0) \\ C&=(1,0) \\ \bigcirc ABCD&:\;\;x^2+y^2=1 \\ \bigcirc OBD&:\;\;(x-1)^2+y^2=1 \\ B&=(1,-\sqrt3)/2 \\ D&=(1,\sqrt3)/2 \\ P&=(2t^2,2t)/(t^2+1) \\ CP&:\;\;2t\,x+(1-t^2)\,y=2t \\ BP&:\;\;(t+\sqrt3)\,x+(1-\sqrt3t)\,y=2t \\ A&=(-t^4 + 6t^2 - 1, -4t^3 + 4t)/(t^4 + 2t^2 + 1) \\ AD&:\;\;(1-t^2-2\sqrt3t)\,x+(\sqrt3t^2 - 2t - \sqrt3)y = t^2 - 2\sqrt3t - 1 \\ E&=(-3\sqrt3t^3 + 11t^2 + \sqrt3t - 1, -3t^3 - 3\sqrt3t^2 + 9t + \sqrt3)/(4t^2 + 4) \\ F&=(t^2 + 2\sqrt3t - 1, -\sqrt3t^2 + 2t + \sqrt3)/(2t^2 + 2) \\ \lVert P-F\rVert^2&=(3t^2 - 2\sqrt3t + 1)/(t^2 + 1) \\ \lVert E-F\rVert^2&=(9t^4 - 12\sqrt3t^3 + 18t^2 - 4\sqrt3t + 1)/(4t^2 + 4) \\ \lVert B-F\rVert^2&=4/(t^2 + 1) \\ \lVert P-F\rVert^4 &= \lVert E-F\rVert^2\cdot\lVert B-F\rVert^2 = (\sqrt3t - 1)^4/(t^2+1)^2 \end{align*}

I did all the computation using projective geometry and homogeneous coordinates. So the things I write as $(x,y)/z$ were internally just homogeneous coordinate vectors $(x,y,z)$ in my computation. And of course I used a computer algebra system (sage in this case) to handle all these polynomials.