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Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

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#### Solution

“In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.”

**Proof:** Let ABC be a right triangle where ∠B = 90°.

It has to be proved that AC^{2} = AB^{2} + BC^{2}

**Construction:** Draw BD ⊥ AC

In ΔADB and ΔABC,

∠ADB = ∠ABC [Each is right angle]

∠BAD = ∠BAC [Common angle]

Therefore, by AA similarity criterion, ΔADB ∼ ΔABC

**∴** `("AD")/("AB") = ("AB")/("AC")` .....[Sides are proportional in similar triangles]

⇒ AD x AC = AB^{2} ...(1)

**Similarly, it can be proved that ΔBDC ∼ ΔABC**

**∴** `("CD")/("BC") = ("BC")/("AC")`

⇒ AC x CD = BC^{2} ...(2)

**Adding equations (1) and (2), we obtain**

AB^{2} + BC^{2} = AD × AC + AC × CD

⇒ AB^{2} + BC^{2 }= AC (AD + CD)

⇒ AB^{2} + BC^{2} = AC × AC

⇒ AB^{2} + BC^{2} = AC^{2}

This proves the Pythagoras Theorem.

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