Question

$BC,AB$ and $AC$ are tangents to the circle at $D,E$ and $F$ respectively. $\angle EBD=x^{\circ },\angle FCD=y^{\circ }.$ Then $\angle EDF$ equals

• A. $x+y$
• B. $\frac{x}{2}-y$
• C. ${90}^{\circ }-(x+y)$
• D. $\frac{x+y}{2}$

Answer 1

The correct option is D $\frac{x+y}{2}$

In $△EBD$, we have
$EB=ED$ ....(Tangents from same point are equal)
Hence, $\angle BED=\angle BDE$ ....(isosceles triangle property)
Now, Sum of angles $={180}^o$
$\Rightarrow \angle BED+\angle BDE+x=180$
$\Rightarrow 2\angle BDE=180-x$
$\Rightarrow \angle BDE=90-\frac{x}{2}$
Similarly, $\angle FDC=90-\frac{y}{2}$
Now, $\angle EDB+\angle EDF+\angle FDC=180$
$\Rightarrow 90-\frac{x}{2}+\angle EDF+90-\frac{y}{2}=180$
$\Rightarrow \angle EDF=\frac{x+y}{2}$