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in Geometry by (47.4k points)
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Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the others two sides.

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Best answer

Proof:

We are given a right triangle ABC right angled at B.

We need to prove that AC2 = AB2 + BC2

Let us draw BD ⊥ AC

Now, ΔADB ~ ΔABC

AD/AB = AB/AC (sides are proportional)

AD.AC = AB2 … (1)

Also, ΔBDC ~ ΔABC

CD/BC = BC/AC

CD.AC = BC2 … (2)

Adding (1) and (2)

AD.AC + CD.AC = AB2 + BC2

AC(AD + CD) = AB2 + BC2

AC.AC = AB2 + BC2

AC2 = AB2 + BC2

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