Let ABC be a triangle. The external and internal angle bisectors of ∠CAB intersect side BC at D and E, respectively. Let F be a point on the segment BC. The circumcircle of triangle ADF intersects AB and AC at I and J, respectively. Let N be the mid-point of IJ and H the foot of E on DN. Prove that E is the incenter of triangle AHF, or the center of the excircle.
Proposed by Steve Dinh
Let EA meet circumcircle of ADF at S. our solution will have some steps.
step1 : S,N,H are collinear.
AD is bisector so IDJ is isosceles and DN is perpendicular bisector of IJ so it passes through center. we have ∠DAS = 90 so DN passes through S.
step2 : ESFH is cyclic.
∠FSH = ∠FAD = 90 - ∠FAS = 90 - ∠FDS = 90 - ∠HDE = ∠HEF so ESFH is cyclic.
step3 : E is center of excircle of AHF.
note that instead we can prove D is incenter of AHF.
∠AFD = ∠ASD = ∠EFH ---> DF is angle bisector of ∠AFH.
∠FHD = ∠FES = ∠DHA ---> DH is angle bisector of ∠FHA.
∠DAF = ∠HSF = ∠HED = ∠HAD ---> HA is angle bisector of ∠HAF.
we're Done.
Let $ABC$ be a triangle. The external and internal angle bisectors of $\angle CAB$ intersect side $BC$ at $D$ and $E$, respectively. Let $F$ be a point on the segment $BC$. The circumcircle of triangle $ADF$ intersects $AB$ and $AC$ at $I$ and $J$, respectively. Let $N$ be the mid-point of $IJ $ and $H$ the foot of $E$ on $DN$. Prove that $E$ is the incenter of triangle $AHF$, or the center of the excircle.
Proposed by Steve Dinh