What's the measure of the segment $BD$ in the figure below?

For reference: The angle $B$ of a triangle $ABC$ measures $120^o$. On $AB$, $BC$ and $AC$ points $M$ are marked; $N$ and $L$ respectively. The circumscribed circles to the triangles $AML$ and $LNC$ intersect in $P$. If $BM = 4, BN = 2$ and $MBP=60^o$. Calculate $BP$ (Answer:$6$)

My progress:

Just use CHADU's theorem..

$\therefore BP = BM + BN\\ \boxed{BP = 2 + 4 = 6}$

t remains to be demonstrated that the triangle MNP is equilateral to use the theorem (the geogebra confirms) but I couldn't...

Solution 1:

The angles subtended by segment $ML$ at point $A$ and at point $P$ are equal. The angles subtended by segment $NL$ at point $C$ and at point $P$ are equal.

So show that $\angle MPN = \angle A + \angle C = 60^0$

Then $BMPN$ is cyclic. So $\angle MNP = \angle MBP = 60^0$.

So $\triangle MNP$ is equilateral.

You can also apply Ptolemy's theorem in quadrilateral $BMPN$ to find $BP$.

Solution 2:

This is a straightforward application of Miquel's theorem :

Choose three arbitrary points $M,N,L$ on sides $AB,BC,CA$ of $\triangle ABC$. Then circumcircles of $\triangle$s $AML$, $BMN$ and $CNL$ intersect in a single point (known as Miquel point).

Since circumcircles of $\triangle$s $AML, CNL$ intersect in $P$, circumcircle of $\triangle BMN$ also passes through $P$. Changing the view slightly, $B$ lies on circumcircle of $\triangle MNP$. With some easy angle chasing, $\triangle MNP$ can be shown to be equilateral.

Now applying your mentioned theorem, or as a corollary of Ptolemy's theorem, it follows that $$BP=BM+BN=6$$