Pythagorean

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Statement

• Pythagorean

• Theorem

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In a right triangle

• The sum of the areas of the squares on its
  sides equals the area of the square on its
  hypotenuse

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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.

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Analytical Proof

• We will now discuss
• a proof based on the
• congruence of triangles.

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Consider a right triangle ABC

• ? ABC is right angled
• at C

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It is given that triangle ABC is right angled at
C

• We have to prove that
• Area sq. ACGF
• Area sq. BCHK
• Area sq. ABDE

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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,
• ? ABF ? AEC (i)

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But, the area of the ?ABF is half the area
of the square ACGF.

• ? 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 ? ABF
• ½ area sq. ACGF

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The area of the ?AEC equals half the area of
the rectangle AELM.

• On the other hand,
• ?AEC 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 ?AEC
• ½ area rectangle AELM
• (iii)

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The area of the square ACGF the area of the
rectangle AELM

• We have
• From (i), ? ABF ? AEC
• From (ii), area of ? ABF
• ½ area of sq. ACGF
• And from (iii), area of ? AEC
• ½ area of rect. AELM
• Thus, from (i), (ii) and (iii),
• area of sq. ACGF
• area of rect. AELM
• (a)

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In the same way,

• Can you establish that
• The area of sq. BKHC
• area of rect. BDLM?

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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,
• ? ABK ? DBC
• (iv)

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The area of the ?ABK equals half the area of
the square BKHC

• ? ABK has base BK
• The altitude from A BC.
• Therefore,
• area of ? ABK ½ area of square BKHC
• (v)

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The area of the ?BDC equals half the area of
the rectangle BDLM.

• On the other hand,
• ? BDC has base BD
• The altitude from C BM,
• Therefore,
• the area of ? BDC
• ½ area of rect. BDLM
• (vi)

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Thus, the area of the square on side BC equals
the area of the rectangle BDLM

• We now have
• ? ABK ? DBC (iv)
• area of ?ABK
• half area of sq. BKHC.. (v)
• And area of ?DBC
• half area of rect. BDLM.. (vi)
• From (iv), (v) and (vi),
• area of sq. BKHC
• area of rect. BDLM
• (b)

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Combining the results

• area of sq. ACGF area of rect. AELM (a)
• And also,
• area of sq. BKHC area of rect. BDLM (b)

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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

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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

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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
• !!!!

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Verification

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

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