[Aaron.Graeve]

Edit: Incorrect, please disregard
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I would have responded earlier, but I was at an off-season.

Also, I will not have any pretty MS-Paint pictures due to limited computer access. Please bear with me.

The 3 things we know are:
1. OL is perpendicular to DH.
2. DE is perpendicular to OC
3. DE = CH.

Based on 1 and 3, I can safely state that triangles OLC and OLD are similar right triangles reflected along line OL. This is known because both triangles share a leg, a right angle, a hypotenuse, and a terminating point O.

I am given DR, DL, OL, and DE/DF.

From triangles OLC and OLD's similarity, I know LC and DL are the same length. I can also compute OD (which is equal to OC) from the Pythagorean theorem using OL and CL

I know the angle PCL = OCL and that angle can be found by taking arctan(OL/LC)

I can now find PD based on the law of sines. For all of you who have forgotten geometry, the length of a side in a triangle devided by the sine of the angle opposing it js a constant for the triangle. All three side/and pairs share this property. CD/sin(CPD) = PD/sin(PCD). We know both angles and length CD (angle PCD is known from arctan(OL/CL).

Now that I have established PD, I will try and find ED. I know the total angle of COR is arctan(CL/OL) + arctan(RL/OL). OP can be found by using the Pythagorean theorem on the triangle OPD as both PD and OD are known and angle OPD is right by 2.

Angle COR is also and POF. Using triangle POF is right by 2, PF = OP×tan(POF). As we know know PF and PD, DF = PD-PF. DE = DE/DF (given) × DF.

EP = DE-PD. We know the angle HOL is equal to arctan(EP/OP)+arctan(PD/OP)-arctan(DL/OL). We also know triangle HOL is right by 1.

DH =tan(HOL)×OL.

If I have made a mistake, please let me know. I think my solution is correct, but I may be wrong.