Question

State and prove Pythagoras' theorem

Answer 1

Step 1: State the law:

Pythagoras' theorem states that,
“In a right-angle triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides”

In $△ABC$,
$A{C}^2=A{B}^2+B{C}^2$

Step 2: Prove the law

Let $△ABC$ be a right-angle triangle at $\angle B$

In $△CDB$ and $△ABC$
$\angle C=\angle C$ (common)
$\angle CDB=\angle ABC=90°$
Thus, $△CDB~△CBA$

So,
$\begin{array}{cc}& \frac{CD}{BC}=\frac{BC}{AC}\\ \Rightarrow & B{C}^2=CD·AC\dots \left(1\right)\end{array}$

In $△ADB$ and $△ABC$,
$\angle A=\angle A$ (common)
$\angle ADB=\angle CDB=90°$
Thus, $△ADB~△ABC$

So,
$\begin{array}{cc}& \frac{AC}{AB}=\frac{AB}{AD}\\ \Rightarrow & A{B}^2=AC·AD\dots \left(2\right)\end{array}$

Adding $\left(1\right)$ and $\left(2\right)$,
$\begin{array}{cc}& A{B}^2+B{C}^2=AC·AD+CD·AC\\ \Rightarrow & A{B}^2+B{C}^2=AC\left(AD+CD\right)\\ \Rightarrow & A{B}^2+B{C}^2=A{C}^2\mathbf{}\left(\because AD+CD=AC\right)\end{array}$

Hence proved