In ∆ABC, if BD ⊥ AC and BC2 = 2 AC .

In $\triangle \mathrm{ABC}$, if $\mathrm{BD} \perp \mathrm{AC}$ and $\mathrm{BC}^{2}=2 \mathrm{AC}$. $\mathrm{CD}$, then prove that $\mathrm{AB}=\mathrm{AC}$.


Since $\triangle A D B$ is right triangle right angled at $D$

$A B^{2}=A D^{2}+B D^{2}$

In right $\triangle B D C$, we have

$C D^{2}+B D^{2}=B C^{2}$

Since $2 A C \cdot D C=B C^{2}$

$\Rightarrow D C^{2}+B D^{2}=2 A C . D C$

$2 A C \cdot D C=A C^{2}-A C^{2}+D C^{2}+B D^{2}$

$A C^{2}=A C^{2}+D C^{2}-2 A C . D C+B D^{2}$

$A C^{2}=(A C-D C)^{2}+B D^{2}$

$A C^{2}=A D^{2}+B D^{2}$

Now substitute $A D^{2}+B D^{2}=A B^{2}$

$A C^{2}=A B^{2}$

$A C=A B$



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