MathItems

Proof P41 of T39

Right-angled triangle with hypotenuse c, legs a and b, and height h

The area of the triangle is expressible in two ways:

12ab=12ch  ,\tfrac{1}{2} a b = \tfrac{1}{2} c h \; ,

leading to

c=abh  .c = \frac{a b}{h} \; .

Squaring both sides and using the Pythagorean theorem, we obtain

a2b2h2=c2=a2+b2  .\frac{a^2 b^2}{h^2} = c^2 = a^2 + b^2 \; .

Dividing both sides by a2b2a^2 b^2, we get

1h2=a2a2b2+b2a2b2=1a2+1b2  .\frac{1}{h^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{a^2} + \frac{1}{b^2} \; .