Question

In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides. Prove it

Answer 1

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Given:- A right triangle ABC right angled at B.

To prove:- $A{C}^2=A{B}^2+B{C}^2$

Construction:- Draw BD⊥AC

Proof:- In △ABC and △ABD

$\angle ABC=\angle ADB$ (Each ${90}^{\circ }$)

$\angle A=\angle A$ (Common)

$\therefore △ABC\sim △ABD$ (By AA similarity)

$\frac{AC}{AB}=\frac{AB}{AD}$ (∵ Sides of similar triangles are proportional)

$\Rightarrow A{B}^2=AD\cdot AC.....\left(1\right)$

Similarly, in $△ABC$ and $△BCD$, we have

$\angle ABC=\angle BDC$ (Each 90°)

$\angle C=\angle C$ (Common)

$\therefore △ABC\sim △BCD$ (By AA)

$\therefore \frac{BC}{DC}=\frac{AC}{BC}$​

$\Rightarrow B{C}^2=DC\cdot AC.....\left(2\right)$

Adding equation (1) & (2), we have

$A{B}^2+B{C}^2=AD\cdot AC+DC.AC$

$\Rightarrow A{B}^2+B{C}^2=AC(AD+DC)$

$\Rightarrow A{B}^2+B{C}^2=A{C}^2$

Hence proved.