# Triangle with a 60 degrees angle

### What Might This Be About?

### Problem

In $\Delta ABC,$ $\angle BAC=60^{\circ}.$ $BE$ and $CF$ are two angle bisectors meeting at the incenter $I,$ $E\in AC,$ $F\in AB.$

Prove that $EI=FI.$

### Proof

$\angle ABC + \angle ACB=120^{\circ},$ implying $\angle EBC + \angle FCB=60^{\circ},$ so that $\angle EIF=\angle BIC=120^{\circ}$ and $\angle EAF + \angle EIF=180^{\circ},$ making quadrilateral $AEIF$ cyclic.

The chords $EI$ and $FI$ are equal as subtending equal inscribed angles.

### Acknowledgment

The applet above illustrates a portion of Leo Giugiuc's result dealt with elsewhere.

[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny