Question

Prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the square of remaining two sides.

Answer 1

Draw perpendicular BD from the vertex B to the side AC. A – D

In right-angled ΔABC

seg BD ⊥ hypotenuse AC.

∴ By the similarity in right-angled triangles

ΔABC ~ ΔADB ~ ΔBDC

Now, ΔABC ~ ΔADB

∴ `"AB"/"AD" = "AC"/"AB"`   ...(c.s.s.t)

∴ AB² = AC × AD   ...(1)

Also, ΔABC ~ ΔBDC

∴ `"BC"/"DC" = "AC"/"BC"`   ...(c.s.s.t)

∴ BC² = AC × DC   ...(2)

From (1) and (2),

AB² + BC² = AC × AD + AC × DC

= AC × (AD + DC)

= AC × AC   ...(A – D – C)

∴ AB² + BC² = AC²

i.e., AC² = AB² + BC²

Answer 2

Given: In ΔPQR, ∠PQR = 90°.

To prove: PR² = PQ² + QR².

Construction:

Draw seg QS ⊥ side PR such that P–S–R.

Proof: In ΔPQR,

∠PQR = 90°   ...(Given)

Seg QS ⊥ hypotenuse PR   ...(Construction)

∴ ΔPSQ ∼ ΔQSR ∼ ΔPQR   ...(Similarity of right-angled triangles)   ...(1)

ΔPSQ ∼ ΔPQR   ...[From (1)]

∴ `"PS"/"PQ" = "PQ"/"PR"`   ...(Corresponding sides of similar triangles are in proportion)

∴ PQ² = PS × PR   ...(2)

ΔQSR ∼ ΔPQR   ...[From (1)]

∴ `"SR"/"QR" = "QR"/"PR"`   ...(Corresponding sides of similar triangles are in proportion)

∴ QR² = SR × PR   ...(3)

Adding (2) and (3), we get

PQ² + QR² = PS × PR + SR × PR

∴ PQ² + QR² = PR(PS + SR)

∴ PQ² + QR² = PR × PR   ...(P–S–R)

∴ PQ² + QR² = PR² OR PR² = PQ² + QR².