Proposition I.47 · Theorem
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.
Drag the points circled in red. The figure changes, and every step of the proof still holds.
- Let ABC be a right-angled triangle having the angle BAC right. The claim: the square on BC is equal to the squares on BA, AC. (Drag A: it slides along the semicircle on BC, so the angle stays right.)
- On BC let the square BDEC be described, and on BA, AC the squares GB, HC.Proposition I.46Proposition I.46: On a given straight line to describe a square.
- Through A let AL be drawn parallel to either BD or CE.Proposition I.31Proposition I.31: Through a given point to draw a straight line parallel to a given straight line.
- Let AD, FC be joined.Postulate 1Postulate 1: To draw a straight line from any point to any point.
- Since each of the angles BAC, BAG is right, with the 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. For the same reason BA is also in a straight line with AH.Definition 22Proposition I.14Definition 22: Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia.Proposition I.14: If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.
- 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.Postulate 4Common Notion 2Postulate 4: That all right angles are equal to one another.Common Notion 2: If equals be added to equals, the wholes are equal.
- Since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively, and the angle ABD is equal to the angle FBC; therefore the base AD is equal to the base FC, and the triangle ABD is equal to the triangle FBC.Proposition I.4Proposition I.4: If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively.
- The parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallels BD, AL. And the square GB is double of the triangle FBC, for they again have the same base FB and are in the same parallels FB, GC.Proposition I.41Proposition I.41: If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.
- But the doubles of equals are equal to one another; therefore the parallelogram BL is also equal to the square GB.
- Similarly, if AE, BK be joined, the triangle ACE can be proved equal to the triangle KCB; the parallelogram CL is double of the first and the square HC double of the second.Proposition I.4Proposition I.41Proposition I.4: If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively.Proposition I.41: If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.
- Therefore the parallelogram CL is also equal to the square HC.
- Therefore the whole square BDEC is equal to the two squares GB, HC. And the square BDEC is described on BC, and the squares GB, HC on BA, AC. Therefore the square on the side BC is equal to the squares on the sides BA, AC.Common Notion 2Common Notion 2: If equals be added to equals, the wholes are equal.
Being what it was required to prove.