Question

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

Answer 1

“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.