Asked in: July 2025 board paperProve that, ‘In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides’.

Answer: Proof of Pythagoras theorem: \(AC^2 = AB^2 + BC^2\)

[Figure: \(\Delta ABC\) with \(\angle ABC = 90^\circ\); seg BD is drawn perpendicular to hypotenuse AC, D on AC.]

Given: In \(\Delta ABC\), \(\angle ABC = 90^\circ\).

To prove: \(AC^2 = AB^2 + BC^2\).

Construction: Draw seg BD \(\perp\) side AC, A – D – C.

Proof: In right-angled \(\Delta ABC\), BD is perpendicular to the hypotenuse, so \(\Delta ABC \sim \Delta ADB \sim \Delta BDC\) (similarity in right-angled triangles).

From \(\Delta ABC \sim \Delta ADB\): \(\dfrac{AB}{AD} = \dfrac{AC}{AB}\) \(\therefore AB^2 = AD \times AC\) ...... (I)

From \(\Delta ABC \sim \Delta BDC\): \(\dfrac{BC}{DC} = \dfrac{AC}{BC}\) \(\therefore BC^2 = DC \times AC\) ...... (II)

Adding (I) and (II): \(AB^2 + BC^2 = AD \times AC + DC \times AC = AC(AD + DC) = AC \times AC\) (A – D – C)

\(\therefore AB^2 + BC^2 = AC^2\). Hence proved.