prove that in a right triangle the square of the hypotenuse is equal to the aum of the squares of the other two sides

pythagoras theorem:

statement:in a right triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.

__construct a right triange right angled at B.__

construction:construct a perpendicular BD on side AC.

given:angleB=90

to prove:AB^{2}+BC^{2}=AC^{2}

proof: in triangle ABD and tri ABC,

angle A=angle A

angle ADB=angleBDC=90

therefore,triADB is similar to triABC.

which implies AD/AB=AB/AC (sides are in proportion)

which implies AD*AC=AB^{2}-------1

IIIly, tri BDC is similar to tri ABC

BC/DC=AB/BC (sides are in proportion)

which implies AC*DC=BC^{2}-----2

__add 1 and 2__

AD.AC=AB^{2}

AD.AC+AC.DC=AB^{2}+BC^{2}

=AC(AD+DC)=AB^{2}+BC^{2}

=AC^{2}=AB^{2}+BC^{2}

Hence proved.