Proposition I-7

Proposition I-7. Given two straight lines constructed on a straight line and meeting in a point, there cannot be constructed on the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively namely each to that which has the same extremity with it.
Proof.

This is impossible.

The point D cannot be found, which was to be proved.

QED

We have on the one hand that angle ACD is greater than angle BCD, and angle BDC is greater than the angle ADC.

Since angles ACD and ADC are equal and the angles BDC and BCD are equal, this is impossible for the part cannot be greater than the whole. (CN-5)

Thus...

Construct the segment CD.

Consider the triangles ACD and BCD.

By construction, both triangles are isosceles.

Let AB be the given segment, with AC and BC constructed from it.

Suppose that there is another point D for which the segments AD and BD are equal to AC and BC, respectively.

[Correspondence:
AC <---> AD and
BD <---> BC]

Suppose that there is another point D for which the segments AD and BD are equal to AC and BC, respectively.

[Correspondence:
AC <---> AD and
BD <---> BC]

Thus the angles ACD and ADC are equal. (Prop I-5)

So also are the angles BDC and BCD.

Thus the angles ACD and ADC are equal. (Prop I-5)

So also are the angles BDC and BCD.