Question

In a $△ABC$, $D$ is the midpoint of the side $BC.DE$ is perpendicular to $AB$ and $DF$ is perpendicular to $AC$. Also, $DE=DF$, Then $\angle B$ =

• A. $\angle E$
• B. $\angle F$
• C. $\angle C$
• D. $\angle A$

Answer 1

Answer: (C)

$\angle C$

Given: $D$ is the midpoint of $BC$

$\therefore BD=DC$

Also, $DE=DF$

$DE$ is perpendicular to $AB$

$\therefore \angle E={90}^o$

$DF$ is perpendicular to $AC$

$\therefore \angle F={90}^o$

If we consider two triangles so formed, $\Delta DEB$ and $\Delta CFD$, then their hypotenuse $BD$ and $CD$ are equal.

$\angle E=\angle F={90}^o$. Also $DE=CF$.

Thus according to R.H.S condition of congruence $\Delta DEB\cong \Delta CFD$

Thus, the corresponding angles $\angle B$ and $\angle C$ are equal, i.e $\angle B=\angle C$