In $Δ A B C, ∠ A C B$ is an abtuse angle. Seg. AD $⊥$ line BC, then show that: $A B^{2} = B C^{2} + A C^{2} + 2 B C . C D$

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Solution

In $△ A D C$
$∠ D = 90^{\circ}$
$A D^{2} + B D^{2} = A C^{2}$
In $△ A D B$
$∠ D = 90^{\circ}$
$A D^{2} + B D^{2} = A B^{2}$
As we know that $B D = B C + C D$
So, $A B^{2} = A d^{2} + (B C + C D)^{2}$
$⟹ A B^{2} = A C^{2} + B C^{2} + 2 B C + C D$
Hence proved.

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