Relating The Orthocenter To The Circumcircle

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We join $A,P$ .

As $\angle ABP = 90^{\circ}$ , $AP$ is the diameter of the circumcircle of $\triangle ABC$ . Let it's centre be $O$ , incidentally the midpoint of $AP$ . Now, $CH \perp AB$ and $PB \perp AB$ . Therefore, $CH \parallel PB$ . If $C'$ be the midpoint of $AB$ then we know that $CH = 2OC'$ . And applying midpoint theorem in $\triangle ABP$ we have $PB = 2OC'$ . Hence $PB = CH$ and $PB \parallel CH$ . Implies, $CHBP$ is a parallelogram.

Thus we have $CP \parallel BH$ . But $BH \perp AC$ . therefore, $PC \perp AC$ . Therefore, $\angle ACP = 90^{\circ}$ . Moreover, as $CHBP$ is a parallelogram, the diagonals bisect each other. Therefore, $N$ , the intersection of $HP$ with $CB$ is the midpoint of $BC$ . Similarly $N$ is the midpoint of $HP \implies HN=NP=8$ . Now, $M$ is the midpoint of $AC$ and $N$ is the midpoint of $CB$ . Therefore, $MN \parallel AB$ . And we had $CH \parallel PB$ . Therefore $HDBP$ is a parallelogram if $D$ is the foot of the perpendicular from $C$ to $AB$ . But $HD \perp AB$ . Hence $HDBP$ is a rectangle, implying $\angle CHP = 90^{\circ}$ , and our job is almost done. Now, in $\triangle CMP$ $\angle MCP = 90^{\circ}$ and $CH \perp MP$ . From similarity of $\triangle CMH$ and $\triangle CHP$ we have $CH^2 = MH\times HP$ .

Therefore, $CH^2 = 5 \times 18 = 80$ .

Now, in right angled $CHN$ , $CN^2 = CH^2 + HN^2= 80+8^2=80+64=144$ $\implies$ $CN=\sqrt{144} = 12$ . Therefore, $BC = 2CN = 2 \times 12 = \boxed{24}$ .

Denote the circumcenter $O$ , $BO\cap (O)=Y$ . Since $YA\perp AB$ , therefore $YA\parallel CH$ . Similarily $YC\parallel BC$ , hence $AYCH$ is a parallelogram $\Rightarrow Y,M,H,P$ are collinear $\angle MPB=\angle ABP=90\Rightarrow MP\parallel AB$ , this means that $H,P$ lie on the midline of $AB$ $\Rightarrow MP$ bisects $BC$ at $N$ , morever by symmetry $BPCH$ is a parallelogram $\Rightarrow HN=PN=8$ . Hence by similarity $CN*CN=PN*YN=8*(5+5+8)=12^2$ $\Rightarrow CN=12\Rightarrow CB=\boxed {24}$ .

Let $H'$ be the reflection of $H$ across $M$ . Furthermore, let $D$ and $E$ be the feet of the altitudes from $A$ and $C$ to their respective sides. Finally, let $N$ be the midpoint of $BC$ .

First, I claim that $BCH'A$ is cyclic. Indeed, note that as $AM=MC$ and $HM=MH'$ , quadrilateral $AHCH'$ is a parallelogram. Therefore $\angle AH'C=\angle AHC=\pi-(\angle ACH+\angle CAH)=\pi-(\tfrac\pi2-\angle C+\tfrac\pi2-\angle A)=\angle A+\angle C=\pi-\angle B,$ establishing the cyclicity.

Now from the condition that $\angle ABP=90^\circ$ we have $\angle ABP=\angle AH'P=\angle MHC$ , so $MH\perp HC$ . Thus $\triangle MHC\sim\triangle ADC$ , so $MH$ is a midline of $AB$ .

Note $HC=\tfrac12(DB)=8$ . Additionally, since $\triangle ADH\sim\triangle CEH\sim\triangle CHN$ , we have $CH\cdot HD=AD\cdot HN=10\cdot 8=80.$ Therefore $CN=\sqrt{CH^2+HN^2}=\sqrt{80+64}=12$ and the requested answer is $BC=2CN=\boxed{24}$ .

It is seen that: DHPB is rectangular. Call I=AH x BC, H is middle point of CD. So that: $BC^{2}-AC^{2} = 156, BC^{2}+AC^{2}-2BCACcos(C)=676,$ $2BC^{2}-2BCACcos(C)=832, BC^{2}-BCACcos(C)=416$ IHDB is quadrilateral inscribed $\Rightarrow BCIC=CHCD.$ $BC^{2}-BCACcos(C)=BC^{2}-BCIC=BC^{2}-CHCD=BC^{2}-CD^{2}/2=416$ According with $BC^2-CD^2=256 \Rightarrow CD^{2}/2=160 \Rightarrow CD^{2}=320 \Rightarrow BC^{2}=16^2+320$ $\Rightarrow BC=\boxed{24}$