euclidean-geometrygeometryplane-geometry
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…
Observe, $$\angle ICE=\angle IBE=90^{\circ}\implies BECI \;\text{is cyclic}. $$ Let $M$ be the midpoint of $IE$ and the center of the circumcircle of quadrilateral $BECI$. We have, $$ IM=EM=CM=AC.$$ Hence, $$\angle CAM=\angle CMI=2\angle CBI=16^{\circ}$$ Therefore, $$\angle ACB=180^{\circ}-\angle A-\angle B= 180^{\circ}-32^{\circ}-16^{\circ}=\boxed{132^{\circ}}$$
Many thanks to @Ivan for identifying the error in the original image, and here is a solution by geometry as per the request of OP.
Extend $\small PH$ to meet $\small BC$ extend at $\small G$. $\angle CGH=53^\circ$.
$\small \therefore\angle BGP=\angle BAP\implies AGBP\ \text{is cyclic}$
As $\small AB=AP$, $\small \angle ABP=63.5^\circ$, and so is $\small \angle AGP$ because they are in the same segment.
Therefore, $\small \triangle AHG$ is a special right triangle with perpendicular sides in the ratio 1:2.
$\small \therefore HG=6$
Similarly, as $\small \triangle CGH$ is a $\small 3:4:5$ right triangle, $\small HC=8$.
Best Answer
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 $BP$.