Proof
Drop a parallel from a point on to a point on ,
parallel to . Triangles and
share the same base and lie between the same parallels (with
, as transversals through I.29, I.37), so they are equal in
area. Applying VI.1 to versus the equal-area
companion triangles gives . The converse runs the
argument in reverse via I.39.
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Full neighborhood
Depends on (6)
- I.29Proposition I.29A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle…
- I.37Proposition I.37Triangles which are on the same base and in the same parallels are equal to one another.
- I.38Proposition I.38Triangles which are on equal bases and in the same parallels are equal to one another.
- I.39Proposition I.39Equal triangles which are on the same base and on the same side are also in the same parallels.
- VI.1Proposition VI.1Triangles and parallelograms which are under the same height are to one another as their bases.
- V.11Proposition V.11Ratios which are the same with the same ratio are also the same with one another.
Required by (dependents) (8)
- VI.3Proposition VI.3If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the…
- VI.4Proposition VI.4In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which…
- VI.9Proposition VI.9From a given straight line to cut off a prescribed part.
- VI.10Proposition VI.10To cut a given uncut straight line similarly to a given cut straight line.
- VI.11Proposition VI.11To two given straight lines to find a third proportional.
- VI.12Proposition VI.12To three given straight lines to find a fourth proportional.
- VI.24Proposition VI.24In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.
- XI.17Proposition XI.17If two straight lines be cut by parallel planes, they will be cut in the same ratios.
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