Prove that in a Right Angle Triangle, the Square of the Hypotenuse is Equal to the Sum of Squares of the Other Two Sides. - Mathematics

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Sum

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 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 + BC= AC (AD + CD)

⇒ AB2 + BC2 = AC × AC

⇒ AB2 + BC2 = AC2

This proves the Pythagoras Theorem.

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2018-2019 (March) Delhi Set 2

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