Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
“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 AC2 = AB2 + BC2
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 = AB2 ...(1)
Similarly, it can be proved that ΔBDC ∼ ΔABC
∴ `("CD")/("BC") = ("BC")/("AC")`
⇒ AC x CD = BC2 ...(2)
Adding equations (1) and (2), we obtain
AB2 + BC2 = AD × AC + AC × CD
⇒ AB2 + BC2 = AC (AD + CD)
⇒ AB2 + BC2 = AC × AC
⇒ AB2 + BC2 = AC2
This proves the Pythagoras Theorem.
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- Right-angled Triangles and Pythagoras Property