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Maths-

General

Easy

Question

In the figure, AD is a median to BC in $straight triangle A B C$ and $A E perpendicular B C text . If end text B C equals a comma C A equals b comma A B equals c comma A D equals p comma A E equals h$ . and DE =
$x text , prove that end text b squared plus c squared equals 2 p squared plus 1 half a squared$

The correct answer is: Hence proved

Solution :-
Aim :- Prove that $b squared plus c squared equals 2 p squared plus 1 half a squared$
Hint :- apply in pythagoras theorem on triangles AEB , AEC and AED Find $b squared$ and $c squared$ values .Now, add them and do necessary modification to get in terms of p and a.
Explanation(proof ) :-
As D is mid point BD = DC = BC/2 = a/2
Applying pythagoras theorem to triangle AEB we get ,
$A C squared equals E C squared plus A E squared not stretchy rightwards double arrow b squared equals left parenthesis x plus a divided by 2 right parenthesis squared plus h squared minus Eq 1$
Applying pythagoras theorem to triangle AEC we get ,
$A B squared equals B E squared plus A E squared not stretchy rightwards double arrow c squared equals left parenthesis x minus a divided by 2 right parenthesis squared plus h squared minus Eq 2$

Applying pythagoras theorem to triangle AED we get ,
$A D squared equals E D squared plus A E squared not stretchy rightwards double arrow p squared equals x squared plus h squared minus Eq 3$
Adding Eq 1 and Eq 2
We get $b squared plus c squared equals left parenthesis x minus a divided by 2 right parenthesis squared plus h squared plus left parenthesis x plus a divided by 2 right parenthesis squared plus h squared$
[By applying $open left parenthesis a minus b right parenthesis squared plus left parenthesis a plus b right parenthesis squared equals 2 open parentheses a squared plus b squared close parentheses close square brackets$
$not stretchy rightwards double arrow b squared plus c squared equals 2 h squared plus 2 open parentheses x squared plus left parenthesis a divided by 2 right parenthesis squared close parentheses$
$b squared plus c squared equals 2 open parentheses h squared plus x squared close parentheses plus 2 open parentheses a squared over 4 close parentheses$
Substitute Eq3 in the above condition
$b squared plus c squared equals 2 open parentheses p squared close parentheses plus 1 half a squared not stretchy rightwards double arrow b squared plus c squared equals 2 p squared plus 1 half a squared$
Hence proved

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