In $U A B C^e B A C = 90^{o}, A B = A C s e g A P o$ side $B C, B - P - C$ . $D$ is any points on side $B C$ . Prove that: $2 A D^{2} = B D^{2} + C D^{2}$

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Solution

$∠ A = 90^{\circ} ∠ B = 45^{\circ} ∠ B A P = 45^{\circ} ⟹ A P = B P \dots (1) ∠ A C P = 45^{\circ} ⟹ A P = P C \dots (2) ⟹ A P = B P = C P$
$⟹ 2 A P^{2} = B P^{2} + C P^{2} 2 (A D^{2} - P D^{2}) = B P^{2} + C P^{2} 2 A D^{2} = B P^{2} + P D^{2} + C P^{2} + P D^{2} 2 A D^{2} = (B P + P D)^{2} - 2 B P . P D + (C P - P D)^{2} + 2 B P . P D 2 A D^{2} = B D^{2} + C D^{2}$

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