To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.

X.31 ad fin.

Set out two medial straight lines *AB* and *BC*, commensurable in square only, such that the rectangle which they contain is rational and the square on *AB* is greater than the square on *BC* by the square on a straight line incommensurable with *AB*.

Describe the semicircle *ADB* on *AB*. Apply a parallelogram to *AB* equal to the square on *BE* and deficient by a square figure, namely the rectangle *AF* by *FB*. Then *AF* is incommensurable in length with *FB*.

Draw *FD* from *F* at right angles to *AB*, and join *AD* and *DB*.

Since *AF* is incommensurable in length with *FB*, therefore the rectangle *BA* by *AF* is also incommensurable with the rectangle *AB* by *BF*.

But the rectangle *BA* by *AF* equals the square on *AD*, and the rectangle *AB* by *BF* equals the square on *DB*, therefore the square on *AD* is also incommensurable with the square on *DB*.

And, since the square on *AB* is medial, therefore the sum of the squares on *AD* and *DB* is also medial.

And, since *BC* is double *DF*, therefore the rectangle *AB* by *BC* is also double the rectangle *AB* by *FD*.

But the rectangle *AB* by *BC* is rational, therefore the rectangle *AB* by *FD* is also rational.

But the rectangle *AB* by *FD* equals the rectangle *AD* by *DB*, so that the rectangle *AD* by *DB* is also rational.

Therefore two straight lines *AD* and *DB* incommensurable in square have been found which make the sum of the squares on them medial, but the rectangle contained by them rational.

Q.E.D.