# Properties of Miquel point

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.

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.