Since the image isn't visible, I'll assume a common geometry problem involving a circle, diameter, and perpendicular chord (a frequent setup):

Problem Assumption:

AB is the diameter of a circle. CD is a chord perpendicular to AB at point E. Given AE = 2 and EB = 8, find the length of CD.

Step-by-Step Solution:

1. Find the radius:
   AB = AE + EB = 2 + 8 = 10.
   Radius ( r = \frac{AB}{2} = 5 ).

2. Locate the center:
   The center O of the circle is the midpoint of AB, so AO = OB = 5.
   Distance from center O to E: ( OE = AO - AE = 5 - 2 = 3 ).

3. Use Pythagoras for half the chord:
   CD ⊥ AB, so the perpendicular from the center to the chord bisects it (circle theorem). Thus, CE = ED.
   In right triangle OEC:
   ( OC^2 = OE^2 + CE^2 )
   ( 5^2 = 3^2 + CE^2 )
   ( CE^2 = 25 - 9 = 16 )
   ( CE = 4 ).

4. Full chord length:
   CD = 2 × CE = 2 × 4 = 8.

Answer: (\boxed{8}) (adjust if the problem differs, but this is a typical result for such setups).

If your problem is different, please describe it, and I’ll help!

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