Properties of Miquel point

Question

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.

Answer 1

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.

Answer 2

I think in the second question you're referring to the circumCENTER of $\triangle QBC$ and $\triangle QAD$,which we denote by $O_3$ and $O_4$ respectively.

For the first part, we use the following lemma:
$M$ is the center of the spiral similarity sending $AB$ to $DC$.
Proof:
since $A,B,P,M$ and $D,C,P,M$ are concyclic, we get that $$\measuredangle AMB=\measuredangle APB=\measuredangle DPC=\measuredangle DMC$$ Similiary $$\measuredangle ABM=\measuredangle DCM$$ and we've proved the lemma.

Now since this spiral similarity sends $AB$ to $DC$,it also sends the circumcenter of $\triangle MAB$ to $\triangle MDC$, namely it sends $O_1$ to $O_2$.
Hence $MO_1O_2$ is sent to $MAD$ by a spiral similarity and this leads to the fact that they're similar.

For the second part we use angle chasing.
Known that $\measuredangle MO_2O_1=\measuredangle MDA$ and $M,Q,D,A$ are concyclic,we can compute $\measuredangle MO_2O_1=\measuredangle AQM=\frac{1}{2}\measuredangle AO_4M$
The last thing we need to notice is that $O_1O_4$ is the perpendicular bisector of $MA$ beacause the latter is the radical axes of two circles with center $O_1$ and $O_4$.
With this fact in mind, we get $\measuredangle O_1O_4M=\frac{1}{2}\measuredangle AO_4M$.
Therefore $\measuredangle MO_2O_1=\measuredangle MO_4O_1$ and these four points are concyclic.Similiarly we can prove that $O_3$ also lies on this circle.

Answer 3

(a) By Steiner-Miquel's complete quadrilateral theorem we know that, circumcircles of the triangles $PAB, PCD, QBC, QAD$ concur at a point $M$ as called Miquel point.

By inscribed angles and center angles, $\angle MO_1A = 2 \angle MPA = 2 \angle MPD = \angle MO_2D$. Since $O_1M = O_1A$ and $O_2M=O_2D$; we find that $MO_1A \sim MO_2D$. By spiral similarity, we yield $MO_1O_2 \sim MAD$.

(b) Since $O_2M = O_2D$, $O_4M =O_4D$; $O_2MO_4D$ is a kite. Similarly $O_1MO_4A$ is a kite. $$ \angle MO_2O_4 = \dfrac{1}{2}\angle MO_2D = \angle MPD = \angle MPA = \dfrac{1}{2}\angle MO_1A = \angle MO_1O_4 .$$

Therefore, the points $M, O_1 , O_2, O_4$ are concyclic. With similar arguments, $O_2MO_3C$ and $MO_1PO_2$ are kites. So,

$$ \angle MO_3O_2 = \dfrac{1}{2}\angle MO_3C = \angle MBC = \angle MBP = \dfrac{1}{2}\angle MO_1P = \angle MO_1O_2 .$$

Thus, the points $M, O_1 , O_2, O_3$ are concyclic. We conclude that $M, O_1 , O_2, O_3, O_4$ are concyclic.