An elegant proof using reflections

[@l4fourier75]

Really beautiful! Especially the explanations how to deal with reflections in the beginning have given me enlighting insight in this scenario. It seems Q to be the incenter, but this does not have to be this way - nevertheless a super useful hint when searching for the solution.

[@Maths_3.1415]

I think this is the shortest video on the channel

[@EgeSaribas05]

There is another proof to this beautiful problem: Let O be the center and r be the radius of (AFE). So: BO² - r² = BF. BA = BD.BC CO² - r² = CE. CA = CD.CB We can substract them: BO² - CO² = BD² - CD² That means OD is perpendecular to BC. So: BO² = BD² + OD² We can use first equation: BD² + OD² - r² = BD.BC Hence: OD² - r² = BD.DC Also if we define T as the second intersection of AD and (AFE): OD² - r² = DT.DA Hence: DP.DA = DB.DC = DT.DA That means DP = DT. Also we know that the symmetry of T whit respect to OD is on the (AFE) again. Let's name that point Q'. By symmetry, DP = DT = DQ'. That means

[@calimath6701]

Cool solution!