1.
Different Proofs of Pythagoras
Theorem
By Ansh,Prakul and Kushagra

2.
Proof 1
• ΔABF=ΔAECbySAS.Thisisbecause,AE=
AB,AF=AC,and
• ∠BAF=∠BAC+∠CAF=∠CAB+∠BAE=
∠CAE.
• ΔABFhasbaseAFandthealtitudefromB
equaltoAC.Itsareathereforeequalshalf
thatofsquareonthesideAC.Ontheother
hand,ΔAEChasAEandthealtitudefromC
equaltoAM,whereMisthepointof
intersectionofABwiththelineCLparallel
toAE.Thus,theareaofΔAECequalshalf
thatoftherectangleAELM.Whichsaysthat
theareaAC²ofthesquareonsideAC
equalstheareaoftherectangleAELM.
• Similarly,theareaBC²ofthesquareon
sideBCequalsthatofrectangleBMLD.
Finally,thetwo rectanglesAELMandBMLD
makeupthesquareonthehypotenuseAB.

3.
Proof 2
• Weknow,△ADB~△ABC
• Therefore,AD/AB=AB/AC(correspondingsidesofsimilartriangles)
• Or,AB² =AD×AC……………………………..……..(1)
• Also,△BDC~△ABC
• Therefore,CD/BC=BC/AC(correspondingsidesofsimilartriangles)
• Or,BC² =CD×AC……………………………………..(2)
• Addingtheequations(1)and(2)weget,
• AB² +BC² =AD×AC+CD× AC
• AB² +BC² =AC(AD+CD)
• Since,AD+CD=AC
• Therefore,AC² =AB² +BC²
• Hence,thePythagoreantheoremisproved

4.
Proof 3
• Westartwithfourcopiesofthesametriangle.Threeof
thesehavebeenrotated90°,180°,and270°,
respectively.Eachhasareaab/2.Let'sputthem
togetherwithoutadditionalrotationssothattheyforma
squarewithsidec.Thesquarehasasquareholewiththe
side(a-b).Summingupitsarea(a-b)² and2ab,the
areaofthefourtriangles(4·ab/2),weget
C² =(a-b)²+2ab
=a²-2ab+b²+2ab
=a²-b²

5.
Proof 4
• Two triangles are said to be similar if their corresponding angles are of
equal measures and their corresponding sides are in the same ratio.
Also, if the angles are of the same measure, then we can say by using
the sine law, that the corresponding sides will also be in the same
ratio. Hence, corresponding angles in similar triangles will lead us to
equal ratios of side lengths.In triangle ABD and triangle ACB:
• ∠A = ∠A (common)
• ∠ADB = ∠ABC (both are right angles)
• Thus, triangle ABD and triangle ACB are equiangular, which means that
they are similar by AA similarity criterion. Similarly, we can prove
triangle BCD similar to triangle ACB. Since triangles ABD and ACB are
similar, we have AD/AB = AB/AC. Thus, we can say that AD × AC = AB2.
Similarly, triangles BCD and ACB are similar. That gives us CD/BC =
BC/AC. Thus, we can also say that CD × AC = BC2. Now, using both of
these similarity equations, we can say that AC2 = AB2 + BC2. Hence
Proved.

6.
Proof 5
• AlgebraicmethodproofofPythagorastheoremwillhelp
usinderivingtheproofofthePythagorasTheoremby
usingthevaluesofa,b,andc(valuesofthemeasuresof
thesidelengthscorrespondingtosidesBC,AC,andAB
respectively).ConsiderfourrighttrianglesABCwherebis
thebase,aistheheightandcisthehypotenuse.Arrange
thesefourcongruentrighttrianglesinthegivensquare,
whosesideisa+b.Theareaofthesquaresoformedby
arrangingthefourtrianglesisc2.Theareaofasquare
withside(a+b)=Areaof4triangles+Areaofsquarewith
sidec.Thisimplies(a+b)2 =4×1/2×(a×b)+c2,a2 +b2 +2ab=
2ab+c2.Therefore,a2 +b2 =c2.HenceProved.

7.
Proof 6
• Thisproofisavariationon#1,one
oftheoriginalEuclid'sproofs.In
parts1,2,and3,thetwosmall
squaresareshearedtowardseach
othersuchthatthetotalshaded
arearemainsunchanged(and
equaltoa²+b².)Inpart3,thelength
oftheverticalportionofthe
shadedarea'sborderisexactlyc
becausethetwoleftovertriangles
arecopiesoftheoriginalone.This
meansonemayslidedownthe
shadedareaasinpart4.Fromhere
thePythagoreanTheoremfollows
easily.