Question

proof of Pythagoras theorem​

Answer 1

Pythagorean theorem, the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)—or, in familiar algebraic notation, a2 + b2 = c2.

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Answer 2

From the above-given figure, consider the ΔABC and ΔADB,

In ΔABC and ΔADB,

∠ABC = ∠ADB = 90°

∠A = ∠A → common

Using the AA criterion for the similarity of triangles,

ΔABC ~ ΔADB

Therefore, AD/AB = AB/AC

⇒ AB2 = AC x AD ……(1)

Considering ΔABC and ΔBDC from the below figure.

Pythagoras Theorem Derivation -3

∠C = ∠C → common

∠CDB = ∠ABC = 90°

Using the Angle Angle(AA) criterion for the similarity of triangles, we conclude that,

ΔBDC ~ ΔABC

Therefore, CD/BC = BC/AC

⇒ BC2 = AC x CD …..(2)

From the similarity of triangles, we conclude that,

∠ADB = ∠CDB = 90°

So if a perpendicular is drawn from the right-angled vertex of a right triangle to the hypotenuse, then the triangles formed on both sides of the perpendicular are similar to each other and also to the whole triangle.

To Prove: AC2 =AB2 +BC2

By adding equation (1) and equation (2), we get:

AB2 + BC2= (AC x AD) + (AC x CD)

AB2 + BC2 = AC (AD + CD) …..(3)

Since AD + CD = AC, substitute this value in equation (3).

AB2 + BC2= AC (AC)

Now, it becomes

AB2+ BC2= AC2

Hence, Pythagoras theorem is proved.