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Prove Pythagoras theorem.

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Proof of the Pythagoras theorem using similarity of triangles.

Given: ∠BAC = 90°

To Prove: BC2 = AB2 + AC2

Construction: Draw AD ⊥ BC

Proof:

In ∆ ABC and ∆ DBA

∠B is common and ∠BAC = ∠ADB = 90°

Therefore, ∆ ABC ~ ∆ DBA, (AA similarity)

Hence, the ratio of corresponding sides are equal.

⇒ ABDB = BCBA

⇒ AB2 = BC x DB

Similarity, ∆ ABC ~ ∆ DAC,

⇒ AC/DC = BC/AC

⇒ AC2 = BC x DB

Adding (1) and (2), we get

AB2 + AC2 = BC x DB + BC x DC

= BC x (DB + DC) = BC x BC

⇒ AB2 + AC2 = BC2 and hence the proof of the theorem.

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