For reference: $\overset{\LARGE{\frown}}{AP} \cong \overset{\LARGE{\frown}}{PC}$
$\overset{\LARGE{\frown}}{PQ}= \overset{\LARGE{\frown}}{AQ}+\overset{\LARGE{\frown}}{BC}$
If $HC=a$ to be calculated $BM$ (Answer: $a\sqrt2$)
My progress
Draw $HM, CQ, CP, PQ$
Th.Ptolemy $BPQC:$
$\boxed{QC.BP = BQ.CP+BC.PQ}$
If $\overset{\LARGE{\frown}}{AP} \cong \overset{\LARGE{\frown}}{PC}$ Can I say $PM$ is perpendicular bissector? $\implies AN = MC?$
TH .Median:
$AB^2+BC^2 = 2BM^2 +\frac{AC^2}{2}$
$BCAP (cyclic):\boxed{BC.AP+AB.CP = AC.BP}\\\triangle HPC:\boxed{CP^2 = a^2+HP^2}\\
HCMP(Cyclic): \boxed{a.PM+CM.PH = CP.HM}$
but but I'm not getting related to the equations???
Original figure:
Best Answer
I found this solution:
$BM=x$
$\angle BAC = \alpha\\ \angle BAM = \theta\\ \therefore \angle NMB = \theta +\alpha\\ \overset{\LARGE{\frown}}{PQ}= \overset{\LARGE{\frown}}{AQ}+\overset{\LARGE{\frown}}{BC}\implies \angle MBN = \theta+\alpha \therefore BN = MN = m\\ \overset{\LARGE{\frown}}{AP}\cong \overset{\LARGE{\frown}}{PC} \implies$
M is midpoint:$AM = MC = m + n (NC = n)$
Power point $N$: $m.PN= n(2m+n) \implies \boxed{PN = \frac{n(2m+n)}{m}}$
Drew $NK \perp MB\\ \angle MNK = (\alpha+\theta) \\ \triangle NMK \rightarrow \boxed{cos(\theta+\alpha) = \frac{x}{2m}}\\ \triangle PCB \sim \triangle NCP \therefore \boxed{PC^2 = PN.PB}\\ PN+m=: \boxed{PB=\frac{(m+n)^2}{m}}\\ \therefore \boxed{PC^2=\frac{n(2m+n)(m+n)^2}{m^2}}\\ \triangle PCH \sim \triangle MNP : \frac{a}{m}=\frac{PC}{PN}$
Replacing PN e PC, and squaring
$\frac{a^2n^2(2m+n)^2}{m^2}=\frac{n(2m+n)(m+n)^2}{m^2}\\ \boxed{a²=\frac{m^2(m+n)^2}{n(2m+n)}}\\ \triangle MNP: cos(2(\theta+\alpha))=\frac{m}{PN}\\ (cos2a =2cos^2a-1) \therefore 2*\frac{x^2}{4m^2}-1=\frac{m^2}{n(2m+n)}\\ \frac{x^2}{2}=1+\frac{m^2}{n(2m+n)} \implies \frac{x^2}{2}=\frac{m²(m+n)²}{n(2m+n)} \\ \therefore x^2 = 2a^2 \implies \boxed{\color{red}x=a\sqrt2} $
(Solution:by jvmago)