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I need the proof of Pythagoras theorem by Leonardo Da Vinci...........

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Here is the answer to your query.

 

 

Given : A right ΔABC right angled at B

To prove : AC2 = AB2 + BC2

 

Construction : Draw AD ⊥ AC

Proof : ΔABD and ΔABC

∠ADB = ∠ABC = 90°

∠BAD = ∠BAC  (common)

∴ ΔADB ∼ ΔABC  (by AA similarly criterion)

⇒ AD × AC = AB2   ...... (1)

 

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA  (common)

∴ ΔBDC ∼ ΔABC  (by AA similarly criterion)

⇒ CD × AC = BC2   ........ (2)

 

Adding (1) and (2) we get

AB2 + BC2 = AD × AC + CD × AC

= AC (AD + CD)

= AC × AC = AC2

∴ AC2 = AB2 + BC2

 

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One of my favourite proofs,

This proof is ascribed to Leonardo da Vinci (1452-1519). Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. (This follows from the observation that the angle ABH is 45. This is so because ABC is right-angled, thus center O of the square ACJI lies on the circle circumscribing triangle ABC. Obviously, angle ABO is 45.) Now,

Area(ABHI) + Area(JHBC) = Area(ADGC) + Area(EDGF). Each sum contains two areas of triangles equal to ABC (IJH or BEF) removing which one obtains the Pythagorean Theorem.

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