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

In△ABC △ACH∠ACB=∠AHC,∠CAB=∠HAC,∠ABC=∠ACHSo△ACH∼△ABCSimilarly△CBH∼△ABC⟹BCAB=BHBCandACAB=AHAC⟹BC2=AB×BH⋯(1)andAC2=AB×AH⋯(2)(1)+(2)⟹BC2+AC2=AB×(BH+AH)=AB2⟹AB2=BC2+CA2Square of hypotenuse is equal to sum of squares of other two sides