Well, I'm not completely sure, but hmm...

@above, how is it obvious from that? Also, what is it isogonal conjugate with respect to?

We require the fact that in a quadrilateral $WXYZ$, internal point $P$ has an isogonal conjugate with respect to it if $\angle WPX + \angle YPZ = 180^{\circ}$ (of course, this can be modified for points outside by using directed angles). Once this is established, it is not difficult to check that $I_A,I_B$ are isogonal in $ABI_CI_D$ with respect to $\angle A,\angle B$, so supposing $\angle BI_AA + \angle I_CI_AI_D = 180^\circ$, then $I_A$'s isogonal conjugate (w.r.t. $ABI_CI_D$) exists and is necessarily $I_B$. And $I_B$'s isogonal conjugate exists as well, so we get $\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}$.

For the sake of completeness... Russia 2017 grade 11 problem 8 wrote: Let $ABCD$ be a given convex quadrilateral. Points $I_A, I_B, I_C, I_D$ are the incenters of triangles $DAB, ABC, BCD, CDA$ respectively. It turned out that $\angle BI_AA+\angle I_CI_AI_D=180^{\circ}$. Prove that $\angle BI_BA+\angle I_CI_BI_D=180^{\circ}$ holds as well. Consider the following result. Claim: For a quadrilateral $ABCD$ and point $P$ in its interior, $\angle APB+\angle CPD=180^{\circ}$ if and only if the isogonal conjugate of $P$ in $ABCD$ exists. (Proof) Let $X, Y, Z, T$ be the projections of $P$ on the sidelines $AB, BC, CD, DA$ respectively. Let $R$ be the intersection of lines $AD$ and $BC$ (working in the projective plane for this...). Let $S$ be the isogonal conjugate of $P$ in triangle $RAB$ and $Q$ the isogonal conjugate of $P$ in triangle $RCD$. Note that $$\angle APB+\angle CPD=180^{\circ} \Longrightarrow \angle PBA+\angle PAB+\angle PCB+\angle PDA=\angle ZWX+\angle ZYX=180^{\circ},$$so $W, X, Y, Z$ are concyclic. Let the circumcircle of this quadrilateral meet lines $AD, BC$ again at $U, V$. Note that $P, S$ share the same pedal circle in triangle $RAB$, so $S$ is the intersection of the perpendiculars from $U, V$ to $AD, BC$ respectively. Same reasoning applies to $Q$ so $S \equiv Q$ is the desired isogonal conjugate. Reversing the previous argument, we can even show the converse. $\square$ To see our original result holds, observe that the condition is equivalent to saying that the isogonal conjugate of $I_A$ in $ABI_CI_D$ exists. Note that $\angle I_AAI_D=\angle I_BAB$ and $\angle I_CBI_B=\angle ABI_A,$ so $I_A, I_B$ indeed are isogonal conjugates. Finally, as the isogonal conjugate of $I_B$ exists, we see that it must satisfy the desired condition. $\square$

Choose $P$ on $AB$ with spiral similarity $\triangle API_B\sim \triangle AI_AI_D$. Since $\angle AI_AI_D+\angle BI_AI_C=\pi$, we have $\triangle BI_AI_C\sim \triangle BPI_B$. By spiral similarity, $\triangle API_A\sim \triangle AI_BI_D, \triangle BPI_A=BI_BI_C$. So $\angle BI_BI_C+\angle AI_BI_D=\angle BPI_A+\angle API_A=\pi$. Done.

what

$\angle BI_AA + \angle I_CI_AI_D=\pi \implies$ $I_A$ has isogonal conjugate wrt $ABI_CI_D$. Since $$\angle I_DAI_A=\angle DAI_A-\angle DAI_D=\angle BAI_B,\text{ } \angle ABI_A=\angle I_CBI_B,$$this isogonal conjugate coincide with $I_B,$ therefore $\angle BI_BA + \angle I_CI_BI_D = \pi.$

\[\text{Easy problem. Diagram not to scale.}\]

Key Claim: $\quad$ If in $ABCD$ there exists a point $X$ s.t. $\measuredangle AXB + \measuredangle CXD = 180^\circ$ then $X$ has an isogonal conjugate $Y$ WRT $ABCD$.

Sketch of Proof

Claim:In $ABI_CI_D$ , $I_A$ and $I_B$ are isogonal conjugates.

Proof

Claim 2 finishes.

It is kinda easy if you know the isogonal conjugate of a point lemma in a quadrilateral(Used in IMO 2018 P6). It boils down to showing that the isogonal conjugate of $I_a$ in $I_cI_DAB$ coincides with $I_B$, which is but a matter of simple angle chasing.

The angle condition means that in $ABI_DI_C$, $I_A$ has an isogonal conjugate. It is easy to sea that $I_B$ is the desired conjugate, say because $\angle DAI_A$ and $\angle I_DAI_B$ are equal as both are half of $\angle BAD$, so $AI_A$, $AI_B$ isogonal, analogous for $BI_B$,$BI_A$.

TestX01 wrote: It is well-known $I_AI_BI_CI_D$ is cyclic. Why?

NO_SQUARES wrote: TestX01 wrote: It is well-known $I_AI_BI_CI_D$ is cyclic. Why? wait nvm i was skibidi capping but it doesn't affect anything