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

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Solution

In right-angled triangle AEB, applying Pythagoras theorem, we have:
$A B^{2} = A E^{2} + E B^{2}$ ...(i)
In right-angled triangle AED, applying Pythagoras theorem, we have:

$A D^{2} = A E^{2} + E D^{2}$
$\Rightarrow A E^{2} = A D^{2} - E D^{2}$ ...(ii)

$Therefore, A B^{2} = A D^{2} - E D^{2} + E B^{2} (from (i) and (ii)) A B^{2} = A D^{2} - E D^{2} + (B D - D E)^{2} = A D^{2} - E D^{2} + (\frac{1}{2} B C^{} - D E)^{2} = A D^{2} - D E^{2} + \frac{1}{4} B C^{2} + D E^{2} - B C . D E = A D^{2} + \frac{1}{4} B C^{2} - B C . D E$

This completes the proof.

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