Question

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

Answer 1

To prove: In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides of the triangle.

Proof: Let PQR be a triangle, right-angled at P.
Draw PS ⊥ QR

Now, we know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

∴ΔQSP ∼ ΔQPR

Therefore, `"QS"/"QP" = "QP"/"QR"` .....................(Since the sides of similar triangles are proportional)

⇒ QS. QR = Qp² .......(1)

Also, we have
ΔPSR ∼ ΔQPR

Therefore, `"RS"/"RP" = "RP"/"RQ"`.............(Since the sides of similar triangles are proportional)

⇒ RS.RQ = RP² … (2)

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

QS.QR + RS.RQ = RP² + QP²

⇒ QR. (QS + RS) = RP² + QP²

⇒ QR.QR = RP² + QP²

⇒ QR² = RP² + QP²

Thus, the square of the hypotenuse is equal to the sum of squares of the other two sides of the triangle.

Answer 2

Given: In ΔABC, m∠B = 90°

To prove: AC² = AB² + BC²

Construction: Draw seg BO ⊥ hypotenuse AC

Proof:

ΔABC ~ ΔADB    ...[similarity in right angled traingles]

∴ `"AB"/"AD" = "AC"/"AB"`    ...[c.s.s.t.]

∴ AB² = AC × AD    ...(I)

ΔABC ~ ΔBDC    ...[similarity in right angled triangles]

∴ `"BC"/"DC" = "AC"/"BC"`    ...[c.s.s.t.]

∴ BC² = AC × DC    ...(II)

Adding (I) and (II) we get,

AB² + BC² = AC × AD + AC × DC

= AC(AD + DC)

= AC × AC    ...[A-D-C]

∴ AB² + BC² = AC²