Denote $CD=d,AB=a,BC=AD=c,CD=b$. Let $h$ be the distance between lines $AB, CD$ and $r_1$ radius of circle inscribed in $\Delta ABC$ and $r_2$ radius of circle exscribed to $\Delta ACD$. It's easy to see that points $IJ$ lie on the same side of line $CD$. Hence we prove that lines $IJ$ and $CD$ are parallel by equality $r_2=h-r_1$which gives the desired conclusion. $r_1=\frac{2[ABC]}{a+c+d}=\frac{ah}{a+c+d}$ $r_2=\frac{2[ACD]}{b+d-c}=\frac{bh}{b+d-c}$ Observe that $ABCD$ is an isosceles trapezoid$\implies \frac{a-b}{2}=\cos\angle ABC\implies ab+c^2=a^2+c^2-2ac\cos\angle ABC=d^2$ after using law of cosines. So $ab=d^2-c^2\iff \frac{a}{a+c+d}=\frac{d-c}{b+d-c}\iff r_1+r_2=h$ QED Greetings from the second side of northern border.

Finally, $\angle AIJ=\angle ACJ=\frac{1}{2} \angle ACD=\frac{1}{2} \angle CAB=\angle IAB$, which implies $IJ\parallel BA.$

Ghoshadi wrote:

Finally, $\angle AIJ=\angle ACJ=\frac{1}{2} \angle ACD=\frac{1}{2} \angle CAB=\angle IAB$, which implies $IJ\parallel BA.$ What exactly do you mean by "angle chasing?"

ProGameXD wrote: Ghoshadi wrote:

Finally, $\angle AIJ=\angle ACJ=\frac{1}{2} \angle ACD=\frac{1}{2} \angle CAB=\angle IAB$, which implies $IJ\parallel BA.$ What exactly do you mean by "angle chasing?" Just click on angle chasing.