Question

Let P be a point inside triangle ABC such that

∠AP B − ∠ACB = ∠AP C − ∠ABC.

Let D, E be the incenters of triangles AP B, AP C, respectively. Show that

AP, BD, CE meet at a point.

Answer 1

Let lines AP, BP, CP

intersect the circumcircle of ∆ABC

again at F, G, H respectively. Now 

∠APB −∠ACB =∠FPG −∠AGB 

 =∠FAG. 

Similarly, ∠APC − ∠ABC = ∠FAH. 

So AF bisects ∠HAG. Let K be the

incenter of ∆HAG. Then K is on AF

and lines HK, GK pass through the 

midpoints I, J of minor arcs AG, AH

respectively. Note lines BD, CE also 

pass through I, J as they bisect ∠ABP, 

∠ACP respectively. H, I on the circumcircle, we see that 

P=BG∩CH, K=GJ∩HI and BI∩CJ= 

BD∩CE are collinear. Hence, BD∩CE is 

on line PK, which is the same as line AP.