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# In the given figure, $ABC$ is triangle in which $\angle ABC > 90^o$ and $AD \perp CB$ produced. Prove that $AC^2 = AB^2 + BC^2 + 2BC \times BD$"

Given:

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

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,

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

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

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

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

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

Hence proved.