Question

In triangle ABC, let AD, BE and CF be the internal angle bisectors with O, E and F on the sides BC, CA and AB respectively. Suppose AD, BE and CF concur at I and B, D, I, Fare concyclic, then $\angle IFD$ has measure

• A. ${15}^o$
• B. ${30}^o$
• C. ${45}^o$
• D. any value $\le {90}^o$

Answer 1

The correct option is B ${30}^o$

$\angle ADB={180}^o-(\frac{A}{2}+B)$
$\angle BFC={180}^o-(\frac{C}{2}+B)$
${180}^o-\frac{A}{2}-B+{180}^o-\frac{C}{2}-B={180}^o$
${180}^o=\frac{A+C}{2}+2B$
$\Rightarrow {360}^o=A+C+4B$
${360}^o=A+B+C+3B$
$\Rightarrow B={60}^o$
$\angle IFD=\angle IBD=\frac{B}{2}={30}^o$