In $\triangle\text{ABC},\text{AB}=\text{AC}.$ Side BC is produced to D. prove that
$(\text{AD}^2-\text{AC}^2)=\text{BD}.\text{CD}$
$(\text{AD}^2-\text{AC}^2)=\text{BD}.\text{CD}$
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Constructing: Draw $\text{AE}\perp\text{BC}.$
Since $\triangle\text{ABC}$ is an isosceles triangle.
We know that in an isosceles triangle,
the altitude and median are the same.
So, $CE = BE$
$\Rightarrow DE + CE = BE + DE = BD$
In $\triangle\text{AED},$
By pythagoras Theorem,
$AD^2 = AE^2 + DE^2$
$\Rightarrow AE^2 = AD^2 - DE^2 ......(i)$
In $\triangle\text{AEC},$
By Pythagoras Theorem,
$AC^2 = AE^2 + EC^2$
$\Rightarrow AE^2 = AC^2 - EC^2 ....(ii)$
From (i) and (ii), we have
$AD^2 - DE^2 = AC - EC^2$
$\Rightarrow AD^2 - AC^2 = DE^2 - EC^2$
$\Rightarrow AD^2 - AC^2 = (DE - EC) (DE + EC)$
$\Rightarrow AD^2 - AC^2 = (CD) (BD)$
$\Rightarrow AD^2 - AC^2 = BD \times CD$
Hence proved.
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