In the figure,$ABC$ is a triangle in which $\angle ABC > 90^{\circ}$ and $AD \perp CB$ produced. Prove that $AC^{2} = AB^{2} + BC^{2} + 2BC \cdot BD$.
(N/A) In $\triangle ADB$,applying the Pythagoras theorem,we have:
$AB^{2} = AD^{2} + BD^{2} \quad \dots(1)$
In $\triangle ADC$,applying the Pythagoras theorem,we have:
$AC^{2} = AD^{2} + DC^{2}$
Since $DC = DB + BC$,we can write:
$AC^{2} = AD^{2} + (DB + BC)^{2}$
$AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB \cdot BC$
Substituting $AD^{2} + DB^{2} = AB^{2}$ from equation $(1)$:
$AC^{2} = AB^{2} + BC^{2} + 2BC \cdot BD$
$AB^{2} = AD^{2} + BD^{2} \quad \dots(1)$
In $\triangle ADC$,applying the Pythagoras theorem,we have:
$AC^{2} = AD^{2} + DC^{2}$
Since $DC = DB + BC$,we can write:
$AC^{2} = AD^{2} + (DB + BC)^{2}$
$AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB \cdot BC$
Substituting $AD^{2} + DB^{2} = AB^{2}$ from equation $(1)$:
$AC^{2} = AB^{2} + BC^{2} + 2BC \cdot BD$
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