Question

Theorem 6.8: In a right triangle, the square of the hypotenuse is equal to the
sum of the squares of the other two sides​

Answer 1

(Diagram is given in attachment)

Given:

• ΔABC is a right Angle Triangle right angled at B

To prove:

AC² = AB² + BC²

Construction:

Draw BD ⟂ AC

Proof:

We know that,

If a perpendicular is drawn from vertex of Right angle in a right angled Triangle to the Hypotenuse then Triangles on both sides of the perpendicular are similar to each other and also to the entire Triangle.

$\therefore \bf{\triangle ADB \sim \triangle ABC}\\\\\bf{\triangle BDC \sim \triangle ABC}\\\\\\\sf{Since \: \triangle ADB \sim \triangle ABC,}\\\\\sf{Its \: corresponding \: sides \: are \: in\:the \: same \: ratio}\\\\\implies \sf{\dfrac{DB}{BC}=\dfrac{AD}{AB} = \dfrac{AB}{AC}}\\\\\\ \sf{On \: cross \: Multiplication,}\\\\\implies \sf{AB^{2} = AD \times AC \quad \longrightarrow [Equation \: 1]}$

$\bf{Similarly \: \triangle BDC \sim \triangle ABC}\\\\\sf{Its \: corresponding \: sides \: are\:in\:the\:same \: ratio}\\\\\\\implies \sf{\dfrac{BD}{AB}=\dfrac{DC}{BC}= \dfrac{BC}{AC}}\\\\\\\implies \sf{BC^{2} = DC \times AC \quad \longrightarrow [Equation \: 2]}$

$\sf{Adding \: Equations \: 1 \: and \: 2;}\\\\\sf{AB^{2} + BC^{2} = (AD\times AC) + (DC \times AC)}\\\\\\Taking \: out \: AC \: as \: a \: common \: factor,\\\\\sf{AB^{2} + BC^{2} = AC(AD + DC)}\\\\\\\implies \sf{AB^{2} + BC^{2} = AC(AC)}\\\\\\\implies \large \bf{AB^{2} + BC^{2} = AC^{2} }\\\\\large \boxed{Hence \: Proved}$