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.
[Desmos graphs can be found here]
Let ABC be a right-angled triangle with angle BAC being a right angle.
We will prove that the square on BC is equal to the squares on AB and AC.
Use Proposition 46 (the construction of a square on a straight line) to construct the square BDEC on the line BC, the square AGFB on the line AB, and the square AHKC on the line AC.
Through the point A draw the line AL that is parallel to BD and CE.
Join AD and CF.
Angles BAC and BAG are each right angles. The straight line BA falls on the two straight lines AC and AG on different sides of it. The adjacent angles are equal to two right angles, so GA is in a straight line with AC by Proposition 14 (if any line and any two lines not on the same side of it form two adjacent angles that are equal to two right angles, then the two lines will be in a straight line with one another).
Also. by the same reasoning, BA is in a straight line with AH.
The right angle DBC is equal to the right angle FBA.
Add angle ABC to each. Therefore angle DBA is equal to angle FBC.
SIde DB is equal to side BC (on the same square), and side AB is equal to side FB.
Comparing triangles ABD and FBC: Sides AB and BD are equal to sides FB and BC and the angles between the two sides are also equal. So, by Proposition 4 (side-angle-side) the base AD is equal to the base FC and the triangle ABD is equal to the triangle FBC.
By Proposition 41 (if a parallelogram and a triangle share the same base and are in the same parallels, then the parallelogram is double the triangle) the parallelogram BL is double the triangle ABD.
Also by Proposition 41, the square AGFB is double the triangle FBC as they share the same base (FB) and are in the same parallels (FB and GC).
Since the parallelogram BL is double the triangle ABD and the square AGFB is double the triangle FBC, and those two triangles are equal to each other, the parallelogram BL is equal to the square AGFB.
Join AE and BK.
In a similar fashion we can prove that the parallelogram CL is equal to the square AHKC.
The whole square BDEC is equal to the parallelograms BL and CL, which are equal to the two squares AGFB and AHKC.
The square BDEC is described on BC, the square AGFB is described on AB, and the square AHKC is described on AC.
Therefore the square on BC (subtending the right angle) is equal to the squares on AB and AC (the sides containing the right angle).
Q.E.D.
This Proposition is equivalent to the Pythagorean theorem. If the lengths of the two legs of a triangle are a and b and the hypotenuse is c, then a^2 + b^2 = c^2. In this Proposition and proof you can actually see the squares instead of working with an abstract formula. There are many proofs of the Pythagorean theorem – I encourage you to look up some of the others.
George
