In a right $Δ A B C$ right-angled at C, if D is the mid-point of BC, prove that $B C^{2} = 4 (A D^{2} - A C^{2})$ .

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Solution

In ACD by Pythagoras theorem,
AD=AC+CD ------------ (i)

In ACB by Pythagoras theorem,
AB= AC+BC --------------(ii)

AD is median, Hence CD = BD = $fraction numerator B C over denominator 2 end fraction$
In eqn (i)
CD=AD - AC
$left parenthesis fraction numerator B C over denominator 2 end fraction right parenthesis squared$ = AD- AC
BC = 4 ( AD - AC)
Hence Proved

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A C^{2} = (4 A D^{2} - 3 A B^{2}) .

ABC is a right angled triangle at C. If D is the mid point of BC,prove that AB²=4AD²-3AC²

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A B^{2} = 4 A D^{2} - 3 A C^{2}

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A C^{2} = 4 A D^{2} - 3 A B^{2}

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