If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right.

In the triangle *ABC* let the square on one side *BC* equal the sum of the squares on the sides *BA* and *AC*

I say that the angle *BAC* is right.

Draw *AD* from the point *A* at right angles to the straight line *AC.* Make *AD* equal to *BA,* and join *DC.*

Since *DA* equals *AB,* therefore the square on *DA* also equals the square on *AB.*

Add the square on *AC* to each. Then the sum of the squares on *DA* and *AC* equals the sum of the squares on *BA* and *AC.*

But the square on *DC* equals the sum of the squares on *DA* and *AC,* for the angle *DAC* is right, and the square on *BC* equals the sum of the squares on *BA* and *AC,* for this is the hypothesis, therefore the square on *DC* equals the square on *BC,* so that the side *DC* also equals *BC.*

Since *DA* equals *AB,* and *AC* is common, the two sides *DA* and *AC* equal the two sides *BA* and *AC,* and the base *DC* equals the base *BC,* therefore the angle *DAC* equals the angle *BAC.* But the angle *DAC* is right, therefore the angle *BAC* is also right.

Therefore *if in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right.*

Q.E.D.