Proof: By Euclid
(related to Proposition: 1.44: Construction of Parallelograms II)
- Let the parallelogram BEFG, equal to the triangle C, have been constructed in the angle EBG, which is equal to D [Prop. 1.42].
- And let it have been placed so that BE is straight-on to AB.
- And let FG have been drawn through to H, and let AH have been drawn through A parallel to either of BG or EF [Prop. 1.31], and let HB have been joined.
- And since the straight line HF falls across the parallels AH and EF, the (sum of the) angles AHF and HFE is thus equal to two right angles [Prop. 1.29].
- Thus, (the sum of) BHG and GFE is less than two right angles.
- And (straight lines) produced to infinity from (internal angles whose sum is) less than two right angles meet together [Post. 5] .
- Thus, being produced, HB and FE will meet together.
- Let them have been produced, and let them meet together at K.
- And let KL have been drawn through point K parallel to either of EA or FH [Prop. 1.31].
- And let HA and GB have been produced to points L and M (respectively).
- Thus, HLKF is a parallelogram, and HK its diagonal.
- And AG and ME (are) parallelograms, and LB and BF the so-called complements, about HK.
- Thus, LB is equal to BF [Prop. 1.43].
- But, BF is equal to triangle C.
- Thus, LB is also equal to C.
- Also, since angle GBE is equal to ABM [Prop. 1.15], but GBE is equal to D, ABM is thus also equal to angle D.
- Thus, the parallelogram LB, equal to the given triangle C, has been applied to the given straight line AB in the angle ABM, which is equal to D.
- (Which is) the very thing it was required to do.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes