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Pythagorean. Theorem. Statement. The sum of the areas of the squares on its sides equals the area of the square on its hypotenuse. In a right triangle . Proofs. There are several proofs of Pythagorean theorem. Some of them are rigorous analytical proofs.

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slide1
Pythagorean

Theorem

Statement

proofs
Proofs

There are several proofs of Pythagorean theorem.

Some of them are rigorous analytical proofs.

They are based on the properties of triangles,

such as congruence of triangles or similarity of triangles.

Some others are based on transformations.

analytical proof
Analytical Proof

We will now discuss

a proof based on the

congruence of triangles.

now consider abf aec they are congruent because
AE = AB (why?)

They are sides of the same

square ABDE

AF = AC, (why?)

They are sides of the same

square ACGF

And also,

BAF = BAC + CAF

= CAB + BAE

= CAE

Hence by SAS,

Δ ABF = Δ AEC …(i)

Now, consider ΔABF & ΔAEC They are congruent, because
but the area of the abf is half the area of the square acgf
Δ ABF has base AF

and the altitude from B

on it = CA

Its area therefore equals half the area of square on the side AF

area Δ ABF

=½ area sq. ACGF

But, the area of the ΔABF is half the area of the square ACGF.
the area of the aec equals half the area of the rectangle aelm
On the other hand,

ΔAEC has base AE

The altitude from C = AM,

(where M is the point of

intersection of AB with

the line CL parallel to AE)

Therefore,

area of ΔAEC

= ½ area rectangle AELM

…(iii)

The area of the ΔAEC equals half the area of the rectangle AELM.
the area of the square acgf the area of the rectangle aelm
We have

From (i), Δ ABF = Δ AEC

From (ii), area of Δ ABF

= ½ area of sq. ACGF

And from (iii), area of Δ AEC

= ½ area of rect. AELM

Thus, from (i), (ii) and (iii),

area of sq. ACGF

=area of rect. AELM

**(a)

The area of the square ACGF = the area of the rectangle AELM
in the same way
Can you establish that

The area of sq. BKHC

=area of rect. BDLM…?

In the same way,
o k let us consider abk dbc they are congruent because
BD = BA (why?)

BC = BK, (why?)

ABK = ABC + CBK

= CBA + ABD

= DBC

Hence by SAS,

Δ ABK = Δ DBC

…(iv)

O.K. -- Let us consider ΔABK & ΔDBC They are congruent, because
the area of the abk equals half the area of the square bkhc
Δ ABK has base BK

The altitude from A = BC.

Therefore,

area of Δ ABK = ½ area of square BKHC

…(v)

The area of the ΔABK equals half the area of the square BKHC
the area of the bdc equals half the area of the rectangle bdlm
On the other hand,

Δ BDC has base BD

The altitude from C = BM,

Therefore,

the area of Δ BDC

= ½ area of rect. BDLM

…(vi)

The area of the ΔBDC equals half the area of the rectangle BDLM.
thus the area of the square on side bc equals the area of the rectangle bdlm
We now have

Δ ABK = Δ DBC … (iv)

area of ΔABK

=half area of sq. BKHC.. (v)

And area of ΔDBC

=half area of rect. BDLM.. (vi)

From (iv), (v) and (vi),

area of sq. BKHC

=area of rect. BDLM

**(b)

Thus, the area of the square on side BC equals the area of the rectangle BDLM
combining the results
area of sq. ACGF =area of rect. AELM **(a)

And also,

area of sq. BKHC =area of rect. BDLM **(b)

Combining the results
summing up
area of sq. ACGF

+area of sq. BKHC

=area of rect. AELM

+area of rect. BDLM

= area of sq. AEDB.

In other words,

area of sq. on side AC

+ area of sq. on side BC.

=

area of the square

on hypotenuse AB

Summing UP
applications
Integers which can form the sides of a right triangle are called Phythagorean Triplets.

Like 3,4 and 5.

And 5,12 and …

???

Think

Calculate

Applications
teaser
(an extension activity)

If two sides AC and BC measure 3 and 4 units -- But if the included angle is not a right angle -- but an obtuse angle, then AB will be…..

More than 5.

Again, if the angle is acute, then

!!!!

Teaser
verification
We will now see a demonstration of verification of Pythagorean theorem through transformations. Verification