Question

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

Answer 1

In $△ABC\&△ACH$

$\angle ACB=\angle AHC,\angle CAB=\angle HAC,\angle ABC=\angle ACH$

So $△ACH\sim △ABC$

Similarly $△CBH\sim △ABC$

$⟹\frac{BC}{AB}=\frac{BH}{BC}$ and $\frac{AC}{AB}=\frac{AH}{AC}$

$⟹B{C}^2=AB\times BH\cdots \left(1\right)$ and $A{C}^2=AB\times AH\cdots \left(2\right)$

$\left(1\right)+\left(2\right)⟹B{C}^2+A{C}^2=AB\times (BH+AH)=A{B}^2$

$⟹A{B}^2=B{C}^2+C{A}^2$

Square of hypotenuse is equal to sum of squares of other two sides