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

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Updated on: 10-Oct-2022

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,$mathrm{AB}^{2}=mathrm{AD}^{2}+mathrm{BD}^{2}$','sharer','toolbar=0,status=0,width=580,height=325');" href="javascript: void(0)" title="Share on Facebook"> 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,$mathrm{AB}^{2}=mathrm{AD}^{2}+mathrm{BD}^{2}$','sharer','toolbar=0,status=0,width=580,height=325');" href="javascript: void(0)" title="Share on LinkedIn">

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