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:AB²+BC²=AC²

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=AB²-------1

IIIly, tri BDC is similar to tri ABC

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

which implies AC*DC=BC²-----2

add 1 and 2

AD.AC=AB²

AD.AC+AC.DC=AB²+BC²

=AC(AD+DC)=AB²+BC²

=AC²=AB²+BC²

Hence proved.

Statement: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

To Prove: c22 = a22 + b22

Proof:

From the above figure, △△ABC is a right angled triangle at angle C.

From C, put a perpendicular to AB at H.

Now, consider the two triangles △△ABC and △△ACH, these two triangles are similar to each other because of AA similarity. This is because both the triangle have a right angle and one common angle at A.

So, by these similarity,

Sum the a22 and b22, we get

acac = eaea and bcbc = dbdb

a22 = c * e and b22 = c * d

a22 + b22 = c * e + c * d

a22 + b22 = c(e + d)

a22 + b22 = c22 (since e + d = c)

Hence Proved.

To Prove- AC²=AB²+BC²

Construction: draw perpendicular BD onto the side AC .

Proof: We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have,
△ADB∼△ABC. (by AA similarity)
 

Therefore, AD/ AB=AB/AC  (In similar Triangles corresponding sides are proportional)

AB²=AD×AC……..(1)

Also, △BDC∼△ABC

Therefore, CD/BC=BC/AC    (in similar Triangles corresponding sides are proportional)

Or, BC²=CD×AC……..(2)

Adding the equations (1) and (2) we get,

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)

( From the figure AD + CD = AC)

AB²+BC²=AC . AC

Therefore, AC²=AB²+BC²

Hope this will help you.....Cheers..