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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

enter image description here

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: enter image description here

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  • 1
    $\begingroup$ Would you mind typing in the problem? I am having problems trying to guess what's given. $\endgroup$
    – Andrei
    Nov 22, 2021 at 17:48
  • $\begingroup$ Can yo please tell me whether BCH is a straight line $\endgroup$
    – creepykrep
    Nov 22, 2021 at 18:02
  • $\begingroup$ @Andrei The reference I post is always an exact copy of the statement $\endgroup$ Nov 22, 2021 at 18:21
  • $\begingroup$ @Bometh yes it's $\endgroup$ Nov 22, 2021 at 18:23
  • $\begingroup$ As the image is not conclusive I now realize that point C in the original image may be the intersection of the circle with the perpendicular... $\endgroup$ Nov 22, 2021 at 18:42

1 Answer 1

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enter image description here

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)

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