Question

$\angle BAC={90}^o$, AD is its bisector. If $DE\perp AC$, prove that $DE\times (AB+AC)=AB\times AC$

• A. $DE\times (AB+AC)=AB\times AC$ is proved
• B. $DE\times (AB+AC)=AB\times AC$is not proved
• C. $DE\times (AB+AC)>AB\times AC$
• D. $DE\times (AB+AC)<AB\times AC$
• E. Blank

Answer 1

The correct option is A $DE\times (AB+AC)=AB\times AC$ is proved

It is given that AD is the bisector of $\angle A$ of $\Delta ABC$.
$\therefore \frac{AB}{AC}=\frac{BD}{DC}$
$\Rightarrow \frac{AB}{AC}+1=\frac{BD}{DC}+1$ [Adding 1 on both sides]
$\Rightarrow \frac{AB+AC}{AC}=\frac{BD+DC}{DC}$
$\Rightarrow \frac{AB+AC}{AC}=\frac{BC}{DC}$........(i)
In ${\Delta }^{\prime }sCDEandCBA$, we have
$\angle DCE=\angle BCA$ [Common]
$\angle DEC=\angle BAC$ [Each equal to ${90}^o$]
So, by AA-criterion of similarity
$\Delta CDE\sim \Delta CBA$
$\Rightarrow \frac{CD}{CB}=\frac{DE}{BA}$
$\Rightarrow \frac{AB}{DE}=\frac{BC}{DC}$.......(ii)
From (i) and (ii), we have
$\Rightarrow \frac{AB+AC}{AC}=\frac{AB}{DE}$
$\Rightarrow DE\times (AB+AC)=AB\times AC$.