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text  IF  end text R text  is a point inside the rectangle  end text A B C D text , prove that  end text RA squared plus R C squared equals R squared plus R D squared

hintHint:

Hint :- Draw perpendicular to both sides of the rectangle .
Now, using pythagoras theorem Find length of RA,RB,RC and RD ,
We know EF is perpendicular to both AB and DC
So we get AE = DF (opposite sides of rectangle AEFD)
EB = FC (opposite sides of rectangle BCFE)

The correct answer is: Hence proved


    Aim  :- Prove that R A squared plus R C squared equals R B squared plus R D squared
    Explanation(proof ) :-
    Draw perpendicular to both sides of the rectangle .
    We know EF is perpendicular to both AB and DC
    It forms two new rectangles
    So we get AE = DF (opposite sides of rectangle AEFD)
    EB = FC (opposite sides of rectangle BCFE)

    Using pythagoras theorem in triangle ARE
    R A squared equals A E squared plus R E squared                             —-Eq1
    Using pythagoras theorem in triangle BRE
    R B squared equals B E squared plus R E squared                              —-Eq2
    Using pythagoras theorem in triangle CRF
    R C squared equals C F squared plus R F squared                              —-Eq3
    Using pythagoras theorem in triangle ARE
    R D squared equals D F squared plus R F squared                              —-Eq4
    Doing Eq1 + Eq3 - (Eq2 +Eq4)
    We get ,R A squared plus R C squared minus open parentheses R B squared plus R D squared close parentheses equals A E squared plus R E squared plus C F squared plus R F squared minus B E squared minus R E squared minus
    D F squared minus R F squared
    Substitute  AE = DF ;EB = FC we get R A squared plus R C squared minus open parentheses R B squared plus R D squared close parentheses equals 0
    Therefore , R A squared plus R C squared equals R B squared plus R D squared
    Hence proved

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