In the given figure, $ABC$ is a triangle in which $\angle ABC = 90^o$ and $AD \perp CB$. Prove that $AC^2 = AB^2 + BC^2 - 2BC \times BD$
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Given:

$ABC$ is a triangle in which $\angle ABC = 90^o$ and $AD \perp CB$.

To do:

We have to prove that $AC^2 = AB^2 + BC^2 - 2BC \times BD$

Solution:

In $\triangle \mathrm{ADB}$,

$\angle \mathrm{ADB}=90^{\circ}$

This implies, by Pythagoras theorem,

$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$.........(i)

In $\triangle \mathrm{ADC}, \angle \mathrm{ADC}=90^{\circ}$

This implies, by Pythagoras theorem,

$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{CD}^{2}$

$=\mathrm{AD}^{2}+(\mathrm{BC}^{2}-\mathrm{BD})^{2}$

$=\mathrm{AD}^{2}+\mathrm{BC}^{2}+\mathrm{BD}^{2}-2 \mathrm{BC} \times \mathrm{BD})$

$=(\mathrm{AD}^{2}+\mathrm{BD}^{2})+\mathrm{BC}^2-2 \mathrm{BC} \times \mathrm{BD}$

$=\mathrm{AB}^{2}+\mathrm{BC}^{2}-2 \mathrm{BC} \times \mathrm{BD}$            [From (i)]

Hence proved.