Distance to Diagonal Intersection

Since the circumcircle of AED is tangent to DC, $\angle EDC=\angle EAD$ . And since EF//DC, so $\angle DEF=\angle EDC=\angle EAD$ . And since $\angle EDF = \angle ADE$ , so triangle EDF is similar to ADE. So $\frac{DA}{DE}=\frac{DE}{DF}$ . So $DE=\sqrt{DA \times DF}=504$ .

Since the circumcircle of triangle $AED$ is tangential to $DC$ , by alternate segments, we know that $\angle CDE = \angle DAE$ . Also, by parallel lines $EF$ and $DC$ , we have that $\angle CDE = \angle FED$ . Hence, $\angle FED = \angle DAE$ . In triangles $DFE$ and $DEA$ , $\angle ADE$ is shared, and thus the condition $\angle FED = \angle DAE$ implies that these 2 triangles are similar by (AAA). Therefore, $\frac{ED}{FD} = \frac{AD}{ED}$ . Simplifying, we solve $ED^2 = (AD)(FD) = (336)(336+420)$ , and $ED = 504$ .

Because FE||DC so then ∠FED=∠EDC

Because DC is tangent at D, ∠FAE=∠EDC (alternate segment theorem)

∴ ∠FED=∠FAE

But for the circum-circle of ΔFAE, ∠FAE is angle in alternate segment to ∠FED and so DE is tangent to this circle at E.

From the rectangle properties of a circle ED²=DF DA=336 756 → ED=504

DE is a chord of the circumcircle \Gamma of triangle AED, and DC is tangent to \Gamma . It follows from the Inscribed Angle Theorem that \angle DAE = \angle CDE.

Now, the triangles AFE and ADC are similar, since EF \parallel CD, and thus \frac{AE}{AC} = \frac{AF}{AD} = \frac{420}{336+420} = \frac{5}{9}. Since CE = CA - AE, we have \frac{CE}{AC} = 1 - \frac{5}{9} = \frac{4}{9}.

Further, the triangles CDE and CAD are similar (since \angle DAE = \angle CDE and \angle DCE = \angle ACD), and hence \frac{AD}{DE} = \frac{AC}{CD} = \frac{CD}{CE} . Put AC = \frac{9}{4}CE to get \frac{9CE}{4CD} = \frac{CD}{CE} , that is, {\frac{CD}{DE}}^{2} = \frac{9}{4}. Hence \frac{CD}{DE} = \frac{3}{2}, and so \frac{AD}{DE} = \frac{3}{2}. Now AD = AF + FD = 336 + 420 = 756, so DE = \frac{2}{3} \times 756 = 504

Let $\Gamma$ be the circumcircle of $ADE$ . Let $G=EF\cap \Gamma$ . Since $DC$ is tangent to $\Gamma$ and $EF$ is parallel to $DC$ , it follows that $GDE$ is isosceles. Thus $\angle GED=\angle DGE=\angle DAE$ . Hence, $ADE\sim EDF$ . It follows that $|ED|^2=|AD|\cdot|FD|\implies ED=504$ .

Let O be the circumcenter of AED, and extend DO into a diameter DR. Let G be the intersection of FE and DO. Since DAR is a right angle, and so is DGF, triangle DAR is similar to DGF. Letting r be the radius |DO|, we have |GD|/|AD| = |DF|/|DR| so |GD| = |AD| |DF| / 2r = (756*336)/2r, and |OG| = r - |GD|.

By Pythagoras on right triangle OGE, |GE|^2 + |OG|^2 = r^2. By Pythagoras on right triangle DGE, |GD|^2 + |GE|^2 = |ED|^2. Hence |ED|^2 = |GD|^2 + r^2 - |OG|^2 = |GD|^2 + (r - |OG|)(r + |OG|) = |GD| ( |GD| + r + |OG| ) = |GD| (2r) = 756*336.

Hence |ED| is the geometric mean of 756 = 84 9 and 336 = 84 4, which is 504 = 84*6.

$\begin{aligned} \angle FED & = \angle EDC & \text{parallel lines}\\ & = \angle EAD & \text{Alternate segment theorem}\\ \end{aligned}$

Hence, triangle $EDF$ and $ADE$ are similar by angle-angle-angle. Thus, $\frac {FD}{ED} = \frac {ED}{DA}$ which gives $ED^2 = FD \cdot DA = 336 \cdot (336 + 420)$ . Hence, $ED = 504$ .

Let circle O be the circumcircle of triangle AED ; and EO be a radius of O . If we assume AD to be the diameter of the circumcircle, then it can be deduced that CD and EF are perpendicular to AD . Because AD is the diameter of circle O , it follows that the radius of circle O is equal to one-half AD or one-half the sum of FA and FD . Therefore, EO =378. Because OD = FD + OF , OF = OD - FD =378-336=42. Using the Pythagorean theorem, EF = \sqrt { EO ^2 - OF ^2 = \sqrt { 378 ^ 2 - 42 ^ 2 }. Using the Pythagorean theorem once more, ED = \sqrt { FD ^2 + EF ^2}= 504