that's embarrassing, but I'm having trouble reading through sentence 1 of Book 10 with Euclid's elements. I struggle with the terminology of Euclid and have no clear idea of what distinctions he makes in the affected lines, so not clear what the evidence says. Here is the text of the proof with some comments / questions from me, embedded in square brackets (I've also numbered the sentences):
Theorem: Let AB and C be two unequal magnitudes, of which AB is the larger one. I say that if an amount greater than half is subtracted from AB, and an amount greater than half is subtracted from AB, and if this process is repeated continuously, an amount remains that remains smaller than the size C.

A multiple DE of C is greater than AB. (Suppose DE is 31 times larger than C and larger than AB, for example.)

Divide DE into parts DF, FG, and GE equal to C. (Euclid can not mean dividing DE into three equal parts of length C. In step 7, he assumes that FD is equal to C, but GE is the same to C?)

Subtract more than half of AB BH, and subtract HK greater than half from AH, and repeat this process until the divisions in AB are equal to most of the divisions in DE. (Does this mean dividing AB into 31 different parts, where BH> half AB and HK> half AH and KL> half AK and so on 31 times?)

Then AK, KH and HB divisions, which correspond in their amount DF, FG and GE. (Does Euclid just say that he considers each line divided into three parts?)

Now that DE is greater than AB and has been subtracted less than half from DE EC (Why is EC less than half of DE?) Do we have to assume here that EG is C? I can not imagine what assumptions we make about the division of DE in step 2), and AB BH is greater than its half, therefore the remainder GD is greater than the remainder HA.

And since GD is greater than HA and GD is half GF (Should GF be half of GD?) And subtracted by HA HK greater than half its size, the remainder DF is larger than the remainder AK.

But DF is equal to C, so C is also bigger than AK. Therefore, AK is smaller than C.

The size AB therefore leaves the size AK which is smaller than the specified smaller size, namely C.