Question

What is the proof of Pythagoras Theorem tell me fast

Answer 1

$<b><i>$

$\huge{ Hey there !}$

Given :- ∆ ABC right angled at B.

To prove :- AC^2= AB^2+ BC^2

Construction :- Draw BD perpendicular on AC.

Proof :-

∆ ADB~ ∆ABC

AD/AB = AB/AC ••••••••••( Sides are propertional )

AD• AC = AB^2 •••••••••(1)

∆ BDC ~ ∆ ABC

CD/ BC = BC/ AC

CD• AC = BC^2 •••••••••(2)

Adding (1) and (2)

AD• AC + CD• AC = AB^2 + BC^2

AC ( AD+ CD ) = AB^2+ BC^2

AC• AC = AB^2+ BC^2

AC^2 = AB^2+ BC^2

Hence proved.......

$<marquee> Hope it helps you !✌✌$

Answer 2

here it is................

In short it is written as: a2 + b2 = c2 



0Save

Let QR = a, RP = b and PQ = c. Now, draw a square WXYZ of side (b + c).  Take points E, F, G, H on sides WX, XY, YZ and ZW respectively such that WE = XF = YG = ZH = b.








0Save

Then, we will get 4 right-angled triangle, hypotenuse of each of them is ‘a’: remaining sides of each of them are band c. Remaining part of the figure is the 

square EFGH, each of whose side is a, so area of the square EFGH is a2.

Now, we are sure that square WXYZ = square EFGH + 4 ∆ GYF

or, (b + c)2 = a2 + 4 ∙ 1/2 b ∙ c

or, b2 + c2 + 2bc = a2 + 2bc

or, b2 + c2 = a2


mark brainliest