In a right $Δ$ ABC 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

$∠ C = 90^{\circ}$
${A B}^{2} = {B C}^{2} + {A C}^{2}$
$= {A C}^{2} + {(2 C D)}^{2}$ [BC=2CD]$
$= {A C}^{2} + 4 {C D}^{2}$ ( $A C D$ is also right triangle ${A D}^{2} = {D C}^{2} + {A C}^{2}$ )
$= {A C}^{2} + 4 ({A D}^{2} - {A C}^{2})$ ( $C D^{2} = A D^{2} - A C^{2}$ )
$= 4 {A D}^{2} - 3 {A C}^{2}$

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Similar questions

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

∠ C = 90^{o}

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

△ A B C

B

D

B C

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

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