Problem #PRU-100260

Problem

In every right-angled triangle the arm is greater than the hypotenuse. Consider a triangle $ABC$ with right angle at $C$.

The difference of the squares of the hypothenuse and one of the arms is $A{B}^2-B{C}^2$. This expression can be represented in the form of a product $$A{B}^2-B{C}^2=(AB-BC)(AB+BC)$$ or $$A{B}^2-B{C}^2=-(BC-AB)(AB+BC)$$ Dividing the right hand sides by the product $-(AB-BC)(AB+BC)$, we obtain the proportion $$\frac{AB+BC}{-(AB+BC)}=\frac{BC-AB}{AB-BC}.$$ Since the positive quantity is greater than the negative one we have $AB+BC>-(AB+BC)$. But then also $BC-AB>AB-BC$, and therefore $2BC>2AB$, or $BC>AB$, i.e. THE ARM IS GREATER THAN THE HYPOTENUSE!