Question

it is given that a right traingle ABC, right angle at B. Prove that AC sq = AB sq+ BC sq​

Answer 1

$in \: it \: hypotenuse \: AC \: and \: the \: two \: sides \: \\ are \: AB \: and \: BC. \\ \\ \: </p><p> according \: to \: the \: Pythagoras \: \\ theorem, \: the \: hypotenuse \: square \\ \: is \: equal \: to \: the \: square \: of \: the \: \\ two \: side \\ </p><p></p><p> so \: in \: this \: question \: AC \: square \: is \\ \: equal \: to \: AB \: square \: + \: BC \: \\ square \: \\ </p><p></p><p> hope \: it \: helps \: you \: \\ </p><p></p><p>please \: mark \: it \: as \: brainliest$

Answer 2

Step-by-step explanation:

the image above:

ABD is a right angled traingle

and AB is the hypotaneous

(AB)^2 = (AD)^2 + (BD)^2

(AB)^2 - (BD)^2 = (AD)^2 ---------(i)

now,

ACD is also a right angled traingle

(AC)^2 = (AD)^2 + (DC)^2

(AC)^2 - (DC)^2 = (AD)^2 ----------(ii)

(i) = (ii)

(AB)^2 - (BD)^2 =(AC)^2 - (DC)^2

(AB)^2 +(DC)^2 = (AC)^2 + (BD)^2