Let $ABCD$ be a quadrilateral with $P=AD \cap BC$ and $Q=AB\cap CD$. Let $M$ be the miquel point of the quadrilateral. Prove the following- $\text{1)}$ If $O_1$ and $O_2$ be the centres of $\triangle PAB$ and $\triangle PDC$ then $MO_1O_2 \sim MAD$. $\text{2)}$ $O_1,O_2$ and the circumcircles of $\triangle QBC$ and the circumcircle of $\triangle QAD$ are concyclic.


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We have the following circles:

  • Brown circle $PAB$ with center $O_1$
  • Red circle $PCD$ with center $O_2$
  • Blue circle $QBC$ with center $O_3$
  • Green circle $QAD$ with center $O_4$

All these circles have one common point $M$ (Miquel).

Let us first show that points $O_1,O_2,O_3,O_4$ are concyclic (part 2 of your problem):

$$O_1O_2\bot MP, \ \ O_1O_4\bot MA\implies\angle O_2O_1O_4+\angle AMP\tag{1}=180^\circ$$

On the other side, quadrialteral AMPB is concyclic and therefore:

$$\angle AMP+\angle B=180^\circ\tag{2}$$

By comparing (1) and (2) you get:

$$\angle O_2O_1O_4=\angle B\tag{3}$$

In a similar way:

$$O_3O_2\bot MC, \ \ O_3O_4\bot MQ\implies\angle O_2O_3O_4+\angle QMC=180^\circ\tag{4}$$

On the other side, quadrialteral QMCB is concyclic and therefore:

$$\angle QMC+\angle B=180^\circ\tag{5}$$

By comparing (4) and (5), we get:

$$\angle O_2O_3O_4=\angle B\tag{6}$$

From (3) and (6) we get that angles above the same segment $O_2O_4$ are equal:

$$\angle O_2O_1O_4=\angle O_2O_3O_4$$

...and therefore quadrilateral $O_1O_2O_4O_3$ is concyclic.

Back to part (1) of your problem. Introduce angle $\angle MAD=\beta$:

$$\angle MAD=\angle MAP=\frac12\angle MO_1P=\angle MO_1O_2=\beta\tag{7}$$

On the other side:

$$\angle MAD=\frac12 \angle DO_4M=\angle MO_4O_2=\beta\tag{8}$$

So angles $\angle MO_1O_2$ and $\angle MO_4O_2$ above the same segment $MO_2$ are equal and therefore quadrialteral $O_1O_2MO_4$ is cyclic. We already proved that $O_1O_2O_4O_3$ is cyclic so points $O_1,O_2,O_3,O_4,M$ all must be on the same circle (not shown in the picture).

Now introduce angle $\angle AMD=\alpha$. We have that $DM\bot O_2O_4$ and $AM\bot O_1O_4$. Consequentially:

$$\angle O_1O_4O_2=\angle AMD=\alpha$$

But quadrualteral $O_1O_2O_3O_4M$ is concyclic and therefore:

$$\angle O_1MO_2=\angle O_1O_4O_2=\alpha$$

So we have proved that:

$$\angle AMD=\angle O_1MO_2=\alpha$$

$$\angle MAD=\angle MO_1O_2=\beta$$

So triangles $\triangle MAD$ and $\triangle MO_1O_2$ have the same angles and consequentially they are similar.


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