Let two straight lines DE, DF be set out containing any angle EDF; let DG be made equal to A, GE equal to B, and further DH equal to C; let GH be joined, and let EF be drawn through E parallel to it. [[I. 31](elem.1.31)]

Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF, therefore, as DG is to GE, so is DH to HF. [[VI. 2](elem.6.2)]

But DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF.

Therefore to the three given straight lines A, B, C a fourth proportional HF has been found. Q. E. F.