Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

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Therefore ADEB is a parallelogram;

therefore AB is equal to DE, and AD to BE. [I. 34]

But AB is equal to AD;

therefore the four straight lines BA, AD, DE, EB are equal to one another; therefore the parallelogram ADEB is equilateral.

I say next that it is also right-angled.

For, since the straight line AD falls upon the parallels
AB, DE,

the angles BAD, ADE are equal to two right angles. [I. 29]

But the angle BAD is right;

therefore the angle ADE is also right.

And in parallelogrammic areas the opposite sides and
angles are equal to one another; [I. 34]

therefore each of the opposite angles ABE, BED is also right. Therefore ADEB is right-angled.

And it was also proved equilateral.

Therefore it is a square; and it is described on the straight line AB.

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Q. E. F.

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Proposition 47.

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In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

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Let ABC be a right-angled triangle having the angle
BAC right;

I say that the square on BC is equal to the squares on BA, AC.

For let there be described on BC the square BDEC,
and on BA, AC the squares GB, HC; [I. 46] through A let AL be drawn parallel to either BD or CE, and let AD, FC be joined.

Then, since each of the angles BAC, BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight lines
AC, AG not lying on the same side make the adjacent angles equal to two right angles;

therefore CA is in a straight line with AG. [I. 14]

For the same reason

BA is also in a straight line with AH.

And, since the angle DBC is equal to the angle FBA: for each is right: let the angle ABC be added to each;

therefore the whole angle DBA is equal to the whole angle FBC. [C.N. 2]