Question

In a $\Delta$ABC, $\angle$BAC $={90}^o$, AD is its bisector. If DE$\perp$AC, then DE$\times$ (AB$+$AC)$=$2AB$\times$AC.

• A. True
• B. False

Answer 1

The correct option is B False

As AD is the bisector of angle A

therefore, $\frac{AB}{AC}=\frac{BD}{CD}$

Adding 1 On both sides

$\therefore \frac{AB}{AC}+1=\frac{BD}{CD}+1$

$⟹\frac{AB+AC}{AC}=\frac{BD+CD}{CD}⟹\frac{AB+AC}{AC}=\frac{BC}{CD}-\left(1\right)$

Also in ${\Delta }^{\prime }s$ CDE and CBA

$\angle ACB=\angle DCE$( common angle)

$\angle CED=\angle CAB$ (each 90)

$\Delta CDE$ ~$\Delta CBA$

$\therefore \frac{CD}{CB}=\frac{DE}{BA}⟹\frac{AB}{DE}=\frac{BC}{CD}-\left(2\right)$

From eqs.(1) and (2) we have,

$\frac{AB+AC}{AC}=\frac{AB}{DE}⟹DE(AB+AC)=AC\times AB$

so Given satement is false.