I think it's an old problem,whereas I didn't know where I saw it.It's said that there's no purely geometrical proof,but i really want to see one,though by trigonometry we need only three times trig-Ceva and some angle chasing.

In triangle $ABC$, let $D$ be a point on $BC$. Let $I_1$ and $I_2$ be the incenters of triangles $ABD$ and $ACD$ respectively. Let $O_1$ and $O_2$ be the circumcenters of triangles $AI_1D$ and $AI_2D$ respectively. Let lines $I_1O_2$ and $I_2O_1$ meet at $P$. Show that $PD\perp BC$.

Let $ D_1 \in DI_2, D_2 \in DI_1 $ be the antipode of $ I_1, I_2 $ in $ \odot (O_1), $ $ \odot (O_2), $ respectively. Note that $$ I_1(I_2, D_2; O_2, \parallel I_2O_2) = -1 = I_2(I_1,D_1;O_1,\parallel I_1O_1 ) \ , $$so $ DP $ passes through the reflection of the incenter of $ \triangle ABC $ in the midpoint of $ I_1I_2 $ $ \Longrightarrow $ $ DP $ is parallel to the Newton line of the complete quadrilateral $ D_1I_1D_2I_2, $ hence $ DA, DP $ are isogonal conjugate WRT $ \angle I_1DI_2 $ $ \Longrightarrow $ $ PD \perp BC. $

Here's a less elegant solution: It suffices to show that $PD$ is isogonal to $AD$ WRT $\angle I_1DI_2 \Leftrightarrow \frac{sin \angle ADI_1}{sin \angle ADI_2} \cdot \frac{sin \angle PI_1D}{sin \angle PI_1I_2} \cdot \frac{sin \angle PI_2I_1}{sin \angle PI_2D} = 1$ Let $X,Y$ be the reflections of $I_1, I_2$ across $O_1, O_2$ respectively. Note that $D,I_2,X$ collinear, $D,I_1,Y$ collinear. $\frac{sin \angle PI_1D}{sin \angle PI_1I_2} = \frac{sin \angle PI_1Y}{sin \angle PI_1I_2} = \frac{I_1I_2}{I_1Y}$ (sine rule in $\triangle I_1I_2Y$ noting that $O_2$ is midpoint of $I_2Y$) Similarly $\frac{sin \angle PI_2I_1}{sin \angle PI_2D} = \frac{I_2X}{I_1I_2}.$ So the equality we have to show is $\frac{sin \angle ADI_1}{sin \angle ADI_2} = \frac{I_1Y}{I_2X}$ We have $\angle AI_1Y = \angle AXI_2, \angle AYI_1 = \angle AI_2X$ (since $AI_2DY, AI_1DX$ cyclic) so $\triangle AI_1Y \sim \triangle AXI_2 \Rightarrow \frac{I_1Y}{I_2X} = \frac{AY}{AI_2}$ But since $\angle AI_2Y = \angle ADI_1, \angle AYI_2 = \angle ADI_2$ we get our desired equality.

quite simple with bash... simply by definition of $O_1$ and $O_2$ we know they are midpoint of arc $AD$ in $ABD$ and $ACD$. Let $X$ be such that $DX \perp BC$. Note that we need to prove $\frac{\sin{XDI_1}}{\sin{XDI_2}} . \frac{\sin{I_1I_2O_1}}{\sin{DI_2O_1}} . \frac{\sin{DI_1O_2}}{\sin{I_2I_1O_2}} = 1$. Let $I_1'$ and $I_2'$ be reflection of $I_1$ and $I_2$ across $O_1$ and $O_2$. Note that $\frac{\sin{I_1I_2O_1}}{\sin{DI_2O_1}} = \frac{\sin{I_1I_2O_1}}{\sin{O_1I_2I_1'}} = \frac{I_2I_1'}{I_2I_1}$ and $\frac{\sin{DI_1O_2}}{\sin{I_2I_1O_2}} = \frac{\sin{O_2I_1I_2'}}{\sin{I_2I_1O_2}} = \frac{I_1I_2}{I_1I_2'}$ so we need to prove $\frac{\sin{XDI_1}}{\sin{XDI_1}} . \frac{I_2I_1'}{I_1I_2'} = 1$. $\frac{\sin{XDI_1}}{\sin{XDI_2}} = \frac{\sin{XDI_1}}{\sin{XDI_2}} = \frac{\sin{ADI_1}}{\sin{ADI_2}} = \frac{\sin{BDI_1}}{\sin{CDI_2}}$ so we need to prove $\frac{\sin{BDI_1}}{\sin{CDI_2}} = \frac{I_1I_2'}{I_2I_1'}$ which is true since $I_1$ and $I_1'$ are incenter and excenter in $ABD$ and $I_2$ and $I_2'$ are incenter and excenter in $ACD$.

It's the same as the question below: \odot (O_1)∩ \odot (O_2)=A and B, C∈\odot (O_1), D∈\odot (O_2), AC⊥AD, DO_1∩CO_2=P. Prove that ∠PAD=∠BAC.