### Pythagorean Theorem from “a new angle”

Lets see the legacy theorem “The Pythagoras Theorem” from a totally new angle, I hope you understood (wink).

Proof-

Distance CD = (AB+AE)/10……………..(Piyush Theorem)

10CD = (a-d) +a

5CD + 5CD = 2a-d

5(BD-BC) + 5(CE-DE) = 2a-d

5BD – 5BC + 5CE – 5DE = 2a-d

5CE – 5BC = 2a-d—– (BD=DE)

5(CE-BC) = 2a-d————————eqn.1

Now,

CE^2 = AE^2 – AC^2

BC^2 = AB^2 – AC^2

CE^2- BC^2 = AE^2 – AB^2

(CE+BC)(CE-BC) = (AE +AB) (AE – AB)

(a +d)(CE – BC) = (a +a –d) (a-a +d)

(a +d)(CE-BC) = (2a-d) d

CE – BC = (2a-d) d/ (a +d) , put value from eqn.1

5(2a –d) d/ (a +d) = (2a-d)

5d/ (a +d) = 1

5d = a +d

5d – d = a

a = 4d

If it is a Right Angle Triangle,

(a +d)^2 = a^2 + (a-d) ^2

(a +d)^2 – (a-d) ^2 = a^2

(a + d + a – d)(a + d –a +d) = a^2

(2a)(2d) = a^2

a = 4d QED.

** **Copyrighted © Piyush Goel (1987)