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Nice question. The figures are similar. and from that, we can divide up DX and OX into 2 parts that have the same ratio as the hypotenuses of the 2 triangles in the figure.

Step One
Proof of similarity.
<LOX = <DOX                         Perpendicular Segments form 2 right angles.
<POL = <BOL                         Given
<BOX = <LOP           

I don't know how you've been told to do this. I will just write steps to show that it is true.
<BOX = 90 - <BOL You are just saying that BOX is the compliment of BOL
<POD = 90 - <POL 

Since <BOL = <POL Then their compliments must be equal. [Compliments of equal angles are equal]. 
 
Therefore the two triangles are similar by AA

Now we can use the similarity of parts.

Step Two
Find OX
10/12 = (13 - OX) / OX  Cross multiply
10 OX = 12(13 - OX) Remove the brackets.
10 OX = 156 - 12 OX  Add 12 OX to both sides.
22 OX = 156 Divide by 22
OX = 156 / 22
OX = 7.091 <<<<< Answer

Check
10/12 = 13 - 7.09/7.09
5/6 = 5.91 / 7.09
0.833333 = 0.833333 So the answer checks.
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