In $Δ A B C, D$ is the midpoint of BC and $A E ⊥ B C$ . If $A C > A B$ , show that $A B^{2} = A D^{2} - B C . D E + \frac{1}{4} B C^{2}$ .

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Solution

Given: $Δ A B C, D$ is the midpoint of BC and $A E ⊥ B C$ and $A C > A B$

Then, BD = CD and $∠ A E D = 90^{o}$

Then, $∠ A D E < 90^{o}$ and $∠ A D C > 90^{o}$

In ΔAED,

$∠ A E D = 90^{o}$

Therefore,
$A D^{2} = A E^{2} + D E^{2}$

⇒ $A E^{2} = (A D^{2} - D E^{2})$ ——- (1)

In ΔAEB,

$∠ A E B = 90^{o}$

Therefore,
$A B^{2} = A E^{2} + B E^{2}$ ——- (2)

Putting the value of $A E^{2}$ from (1) in (2), we get

Therefore,
$A B^{2} = (A D^{2} - D E^{2}) + B E^{2}$

$= (A D^{2} - D E^{2}) + (B D^{2} - D E^{2}) [B u t B D = \frac{1}{2} B C]$

$= A D^{2} - D E 2 + (\frac{1}{2} B C - D E)^{2}$

$= A D^{2} - D E^{2} + \frac{1}{4} B C^{2} + D E^{2} - B C . D E$

$A B^{2} = A D^{2} - B C . D E + \frac{1}{4} B C^{2}$

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