Let ABC be a triangle and D be the foot of perpendicular from A onto BC. Let E and F be incentres of triangles ABD and ADC respectively. The line EF when extended both sides meet AB in K and AC in L. Show that AK = AL if and only if AB = AC or angle A = 90.
I proved that AK = AL when AB = AC. However I am unable to prove it when angle A = 90, also can anyone help me with the proof of the converse. I am unable to conclude anything by angle chasing
It is practical to assume that $D\equiv O$ and set $BD=x_L, DC=x_R$ and $DA=y$. We have:
$$ E = \frac{x_L A + y B}{x_L+y+\sqrt{x_L^2+y^2}},\qquad F = \frac{x_R A + y C}{x_R+y+\sqrt{x_R^2+y^2}}.$$ The constraint $AK=AL$ is equivalent to the orthogonality between the $EF$-line and the angle bisector of $\widehat{BAC}$. Let $N$ be the foot of such angle bisector: by the angle bisector theorem $$ N=\frac{B\sqrt{x_L^2+y^2} + C\sqrt{x_R^2+y^2}}{\sqrt{x_L^2+y^2}+\sqrt{x_R^2+y^2}}$$ and $EF\perp AN$ is equivalent to $$ AE^2-AF^2 = NE^2 - NF^2. $$ I leave to you to compute what this constraint gives in terms of $y,x_L,x_R$, namely $x_L=x_R$ or $x_L x_R = y^2$.