Question

Prove that pythogoras theoram​

Answer 1

Answer:

The proof of Pythagorean Theorem in mathematics is very important. In a right angle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. States that in a right triangle that, the square of a (a2) plus the square of b (b2) is equal to the square of c (c2).

Answer 2

$\Large{\colorbox{black}{\bf{Statement: }}} \: \:$

In a right angle triangle, the square of hypotenuse is equal to the sum of square of other two sides.

$\Large{\colorbox{black}{\bf{Given : }}} \:$

$\sf{\angle{ABC}= 90^\circ} \:$

$\Large{\colorbox{black}{\bf{To Prove: }}} \:$

$\sf{ (AC) ^2 = (AB) ^2+ (BC) ^2} \:$

$\Large{\colorbox{black}{\bf{construction: }}} \: \:$

$\sf{Draw BD \perp \: AC} \:$

$\Large{\colorbox{black}{\bf{Proof: }}} \:$

$\sf{\triangle ADB \sim \triangle ABC} \: \:$

$So, \: \sf \large{\frac{AD}{AB} = \frac {AB}{AC}} \:$

$\fbox{\sf\large{AD \times AC = (AB)^2}\: \rightarrow \: 1}$

Now,

$\sf{\triangle BDC \sim \triangle ABC} \: \: \:$

$So, \: \sf \large{\frac{CD}{BC} = \frac {BC}{AC}} \: \:$

$\fbox{\sf\large{CD \times AC = (BC)^2} \: \rightarrow \: 2 }\:$

Add 1 and 2.,

$\sf{AD \times AC + CD \times AC = (AB) ^2 + (BC) ^2} \:$

$\sf{AC ( AD + CD) = (AB) ^2 + (BC) ^2 } \:$

$\sf{AC( AC) = (AB) ^2+ ( BC) ^2} \:$

$\fbox{\sf{(AC)^2 = (AB) ^2+ ( BC) ^2}}\: \:$

$\huge\mathfrak\red{Hence \: \: Proved} \:$