How to prove that $EO_1FO_2$ is a square?

Question

In triangle $ABC$, $BD$ is the altitude, $O_1$ and $O_2$ are the centers of the incircles $(ADB)$ and $(CDB)$. The circumcircle $(O_1DO _2)$ intersect at points $E$ and $F$. How to prove that $EO_1FO_2$ is a square?

My observations

$\angle O_1DO_2= 45°+45°=90°$

$\angle EO_2F=\angle EDF=90°$

How to prove that $EO_2=EO_1$? for that i need to prove that the smaller triangles are congruent.

Any help would be appreciated.

Answer 1

Inscribed angles of $45^\circ$ at $D$ subtend arcs of $90^\circ$. $\square$

Answer 2

$\triangle O_{1}FO_{2}$ and $\triangle O_{1}EO_{2}$ are both right triangle with $\angle F$ and $\angle E$ being the respective right angles.

Now, $\angle FO_{1}O_{2}=\angle FDO_{2}=45^{\circ}$ and thus $\triangle FO_{1}O_{2}$ is right isosceles triangle.

Similarly, $\triangle O_{1}EO_{2}$ is right isosceles as well and hence $FO_{1}EO_{2}$ is a square.