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#1 h Apr 16, 2006, 1:09 AM • 2 Y

Y by Adventure10, Mango247

An acute triangle $\triangle{ABC}$ is inscribed in a circle with centre $O$ . The altitudes of the triangle are $AD,BE$ and $CF$ . The line $EF$ cut the circumference on $P$ and $Q$ .
a) Show that $OA$ is perpendicular to $PQ$ .
b) If $M$ is the midpoint of $BC$ , show that $AP^2=2AD\cdot{OM}$ .

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Jutaro

#2 h Apr 16, 2006, 5:06 AM • 2 Y

Y by Adventure10, Mango247

Part (a) is very simple. Just observe that $\angle CAO= \pi/2 -\angle ABC$ because $O$ is circumcenter, and $\angle AEF=\angle ABC$ because the quadrilateral $BCEF$ is cyclic. So $\angle CAO+\angle AEF=\pi/2$ , that is, $EF$ , or $PQ$ , is perpendicular to $AO$ .

Next, since $PQ$ is perpendicular to radius $OA$ , this implies by symmetry that $AP=AQ$ . Now we perform an inversion with respect to the circumference of center $A$ and radius $AP$ . The circumcircle of $ABC$ inverts on line $PQ$ , because it passes through the center of inversion. Also, the inverse of $B$ is the intersection of the inverses of line $AB$ and the circumcircle of $ABC$ , that is, the inverse of $B$ is $F$ . In the same way, we see that the inverse of $C$ is $E$ . Now observe that the inverse of line $BC$ is a circumference through $A$ : it must be the circumcircle of $AEF$ , which passes through $H$ , since quadrilateral $AEHF$ is cyclic. But, since $A$ , $H$ and $D$ are collinear, this means that the inverse of $D$ is $H$ . Hence we have $AD \cdot AH =AP^2$ , but we know that $AH=2OM$ , so $AP^2=2AD \cdot OM$ , and we are done

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#3 h Jul 18, 2010, 5:35 AM • 2 Y

Y by Adventure10, Mango247

Part (a) is well known.

As Jutaro said, $AP=AQ$ by symmetry. ( $PO=OQ$ , $AO=AO$ and since $\Delta POQ$ is isosceles and because of part a) we have $<POA=<QOA$ which means that $\Delta AOQ$ is congruent to $\Delta AOP$ so $AP=AQ$ .)

Since cuadrilateral $HECD$ is cyclic then $AH*AD=AE*AC$ . (1)

Since $<BFC=<BEC=90$ cuadrilateral $FECB$ is cyclic and therefore $<AEP=<FEC=180-<FBC=180-<ABC$ (2)

Clearly, cuadrilateral $APCB$ is ciclyc which gives $<APC=180-<ABC$ (3)

From (2) and (3), $<APC=<AEP$ and since $<PAE=PAC$ we have that $\Delta AEP \sim \Delta APC$ and therefore:

$AE*AC= AP^2$ (4)

From (1) and (4), $AH*AD=AE*AC= AP^2$ (5)

It is well known that $AH=2*OM$ and substituting this in 5 we obtain $2*OM*AD= AP^2$ as desired.

[geogebra]df180afdfe95ea51723763af24e90f59689e29e6[/geogebra]

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#4 h Aug 25, 2013, 5:02 PM • 1 Y

Y by Adventure10

Part A: Be K the intersection between prolongation AO and the circuncircle of ABC, <APB=90, is enoguh to show that <APQ=<PKA=b, but <PKA=<ABP=b, <PFB=<ECB=a because FECB is cyclic, PACB is cyclic => <PAB= 180-<ACB => <PAB= 180-a but <PAB= <APQ+<QPB = <APQ + 180-a-b => <APQ + 180-a-b=180-a => <APQ=b => <APK= <PKA=b => AO is perpendicular to PQ.
Part B: as a result in Part A, AP=PQ, by the stewart teorem, AE^2 * QP = AP^2*QP - QE*EP*QP => AE^2=AP^2-QE*EP => AP^2= AE^2+QE*EP but QE*EP=AE*EC by power point, => AP^2=AE^2+AE*EC => AP^2= AE*(AE+EC) => AP^2= AE*AC, but AE*EC= AH*AD by power point, => AP^2=AH*AD, but is known that AH=2OM => AP^2= 2*AD*OM.

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