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Question

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

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Solution

In ABC&ACH
ACB=AHC,CAB=HAC,ABC=ACH

So ACHABC

Similarly CBHABC

BCAB=BHBC and ACAB=AHAC

BC2=AB×BH(1) and AC2=AB×AH(2)

(1)+(2)BC2+AC2=AB×(BH+AH)=AB2
AB2=BC2+CA2

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

1220705_1375149_ans_2301ef3653e14682b35421c5aa81eca5.png

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