AC is tangent to BMP if BC tangent to AMQ

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AC is tangent to BMP if BC tangent to AMQ

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#1 h Oct 7, 2013, 6:04 AM • 2 Y

Y by Adventure10, Mango247

Given an acute angled triangle $ABC$ with $M$ being the mid-point of $AB$ and $P$ and $Q$ are the feet of heights from $A$ to $BC$ and $B$ to $AC$ respectively. Show that if the line $AC$ is tangent to the circumcircle of $BMP$ then the line $BC$ is tangent to the circumcircle of $AMQ$ .

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#2 h Oct 7, 2013, 11:37 AM • 2 Y

Y by Adventure10, Mango247

Without loss of generality, we can assume that $CA<CB$ so that $\odot(BMP)$ is tangent to $AC$ beyond $C$ at $K$ . Then by Power of a point we get $AK=\frac{AB}{\sqrt2}$ and $CK^2=CP.CB$ . But note that $CP.CB$ is also the power of $C$ wrt the circle on diameter $AB$ , hence we get $CP.CB=CM^2-\frac{AB^2}{4}$ . Using the median theorem in a triangle, $CM^2=\frac{2CA^2+2CB^2-AB^2}{4}$ and thus we have $CP.CB=\frac{CA^2+CB^2-AB^2}{2} \implies CK=\sqrt{\frac{CA^2+CB^2-AB^2}{2}}$ .
We then have $AC=AK-CK \implies$ $\frac{AB}{\sqrt2}-AC$ $=\sqrt{\frac{CA^2+CB^2-AB^2}{2}} \implies$ $\frac{AB^2}{2}+AC^2-\sqrt2AB \cdot AC=\frac{CA^2+CB^2-AB^2}{2} \implies 2AB^2-2\sqrt2AC\cdot AB+AC^2-BC^2=0$
Now using the quadratic formula, the length $AB$ is given by $\frac{2 \sqrt2 AC + \sqrt{8AC^2-8(AC^2-BC^2)} }{4}$ (Note that $AB$ is positive so we do not consider the negative root). This gives $AB=\frac{CA+CB}{\sqrt2}$ . Let $L$ be the point on $BC$ such that $BL=AK$ . If $L$ lies beyond $C$ (or $L=C$ ) on $BC$ then we get $\sqrt2AB=CA+CB<AK+BL=\sqrt2AB$ a contradiction. So $L$ is surely between $B,C$ . Thus we get $AK+BL=\sqrt2AB=AC+CB \implies CK=CL$ . Obviously circle $(AML)$ is tangent to $BC$ . if it cuts $CA$ again at $Q'$ then we obtain $CQ'.CA=CL^2=CK^2=CP.CB$ hence, $Q'$ is on $(ABP) \implies Q'$ is the foot of altitude from $B$ thus $Q'=Q$ and $\odot(AMQ)$ is tangent to $BC$ as desired.

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#3 h Oct 7, 2013, 5:36 PM • 2 Y

Y by Adventure10, Mango247

Let $MD$ cut $BC$ at $E$ , easy to see that $\angle EPD = \angle DMB$ and $\angle EDP = \angle PBM = \angle MPB = \angle MDB, \Delta EDP\sim \Delta BMD$ , thus $\angle DEC = \angle DBM = \angle CDE$ , or $EC = DC$ . Using Menelaus in $\Delta ABC$ with line $MED$ easy to obtain $AD = EB$ . Now notice that $BE^{2} = AD^{2} = AM. AB = BM. BA$ , this say that the circle $(AME)$ is tangent to $BE$ . But as $CQ. CA = CP. CB = CD^{2} = CE^{2}, Q$ is a point on this circle and we have q.e.d.

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