Question

In triangle $ABC$, let $AD,BE$ and $CF$ be the internal angle bisectors with $D,E$ and $F$ on the sides $BC,CA$ and $AB$ respectively. Suppose $AD,BE$ and $CF$ concur at $I$ and $B,D,I,F$ are 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 ${30}^o$

Let $\angle IBF=x,\angle BID=y,\angle BIF=z$

$\therefore \angle IBF=IBD=IDF=IFD=x$ using angles in the same arc have the same measure and $BE$ is the angle bisector of $\angle ABC$

Similarly, $\angle BID=\angle BFD=y$

$\angle BIF=\angle BDF=z$

Since points B,D,I,F are con-cyclic, $\angle FBD+\angle FID=180$

$\therefore 2x+y+z=180...\left(1\right)$

In $\Delta FCB,\angle FCB=180-3x-y$

In $\Delta BAD,\angle BAD=180-3x-z$

Summing the angles $A,B,C$ of $\Delta ABC$ to $180$, we have

$2x+2(180-3x-y)+2(180-3x-z)=180$

$\therefore x+180-3x-y+180-3x-z=90$

$\therefore 270=5x+y+z...\left(2\right)$

Subtracting (1) from (2), we have

$90=3x$ or $x={30}^o$