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$M$ is the orthocenter of $\triangle ABC$. $D$ is a point outside of $\triangle ABC$ such that $\widehat{ABD} = \widehat{DBC} = \widehat{DMC}$. $N$ is the orthocenter of $\triangle BCD$. Let $P$ and $Q$ respectively be the circumcenters of $\triangle ABC$ and $\triangle DAB$. Prove that $NP = NQ$.

I tried to let $R$ be the circumcenter of $\triangle BCD$ and prove that $\triangle BRM = \triangle BNQ$. But I still don't know how.

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  • $\begingroup$ I thought the correct statement was that $P=NP$ ... $\endgroup$ – Mirko May 21 at 6:24
  • $\begingroup$ You seem to ask many interesting question like this one, where do you get them from? $MBCD$ is an inscribed quadrilateral if that helps anyhow. $\endgroup$ – Paracosmiste May 21 at 18:43
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Here is another solution which came from "mirror circle"

  1. if circle O is $\triangle ABC $ circum, H is the orthocenter, then the circle O' which pass H,B,C will be the mirror circle of BC with same radius. M is the midpoint of BC, we have $ AH=2OM=OO'$ . and if a $\triangle A‘BC $ is on circle O', then the orthocenter H' must on circle O and $ A'H'=2O'M=OO'$

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since the proof is very simple, I don't give details here.

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solution:

let circle R' is the mirror circle of circle R, connect $RR',OR'$, extend $BD$ to circle R at point $E$, connect $NE$,$NC$,

it is trivial $\angle NEB= \angle NCB $, $N$ is orthocenter $\implies NC\perp BD \implies \angle NEB + \angle DBC =\dfrac{\pi}{2}=\angle NEB+\angle EBA \implies NE \perp AB $

$\angle NDE= \angle BDF, \angle BDF+\angle DBC =\dfrac{\pi}{2} \implies \angle NDE= \angle NED \implies ND=NE=AM=RR'$

$ AB $ on circle $O$ & circle $R \implies OR \perp AB $ with same reason, $OR'\perp BD, RR'\perp BC \implies \angle ROR'= \angle ABD ,\angle RR'O= \angle DBC \implies RR'=OR=NE ,OR // NE \implies NO=ER$

it is trivial $NR=ER$ (radius) $\implies NO=NR$

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  • $\begingroup$ OK, let us compare with "the other proof" (TOP), which takes its time to show that $N$ is on the circle $(ABC)$, and shows the paralelogram $NDMA$. This is 1 above and it is assumed it is so (because it has a name and is simple). Then the proof highly depends on the point $E$, in "the other proof" it is $B^*$. The proof shows first $NE\perp AB$. This is also contained in TOP. The point $R$ is not defined, but i suppose the idea is also to show $NQPE$ parallelogram. Same proof, but using shorter arguments. (Taking time to define all objects makes the proof readable, but slightly longer.) $\endgroup$ – dan_fulea May 24 at 17:18
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Let us draw some more points that may be or not relevant for the solution. (My way to solve synthetic geometry problems is not to restrict to the problem as target, but rather to understand all geometrical relations / connections in the given situation.)

In the following picture...

Problem 3233982, extended picture

the points were constructed in the following order.

  • We start with the $\Delta ABC$.

  • We draw the circumcircle $(ABC)$, and mark with $P$ its center. $P$ was constructed as the intersection of the side bisectors, drawn in green. Among them, the side bisector for the side $AB$ is the more interesting one, since soon we will need the center of an other circle through $A,B$.

  • $M$ is the orthocenter of $\Delta ABC$, the intersection of the heights, and in the picture we also have on two of the heights, the ones from $B,C$, the points $B_1\in AC$, $C_1\in AB$ on the opposite sides, and also $B_2,C_2$, where the heights intersect the circle $(ABC)$.

  • Note that (in measure) $\widehat{B_1MC_1}=\pi-\hat A$, since $\widehat{B_1MC_1}$ and $\hat A=\widehat{BAC}$ are opposite angles in the concyclic $AB_1MC_1$. (Or just drop from $2\pi$ the two right angles in $B_1,C_1$.)

  • Let $BB*$ be the angle bisector of the angle in $B$ in $\Delta ABC$, the point $B^*$ being on the circumscribed circle $(ABC)$ of $\Delta ABC$. The arc $\overset\frown{AC}$ (of length $2\hat B$) is thus cut in two equal parts by $B^*$.

  • Now we construct $D$ as in the problem, insuring that $\widehat DMC$ is half $\hat B$. In particular, $\angle CMD=\angle CBD$ implies $BCDM$ concyclic. This implies also $$ \widehat{BMC}= \widehat{BDC}\ . $$ (So $D$ can be alternatively constructed as the intersection of angle bisector $BB^*$ with the circle $BMC$.)

  • Let now $N$ be the orthocenter of $BDC$. With the same argument as above, showing that the angles $\widehat BAC$ and $\widehat BMC$ add to $\pi$, we also get that $\widehat BDC$ and $\widehat BNC$ add to $\pi$. In particular: $$ \widehat BAC =\pi-\widehat BMC =\pi-\widehat BDC =\widehat BNC\ , $$

  • so $N$ is also on $(ABC)$ and we have $$ PA=PB=PC=PN\ . $$

  • Where on the circle $(ABC)$ is more exactly positioned $N$? (So that we possibly get rid of the definition of $N$ using the complicated point $D$...) We show that $N$ is on the angle bisector of $\angle DCB^*$, and then compute explicitly the angle $\angle NBB^*$, so that $D$ is no longer needed as intermediary station to define $N$. For this, let us compute the angle $\angle NCD$... $$ \begin{aligned} \widehat{NCD} &= \frac\pi2-\widehat{B^*DC} \\ &= \frac\pi2-(\pi-\widehat{BDC}) = \frac\pi2-(\pi-\widehat{BMC}) \\ &= \frac\pi2-\hat A\ ,\qquad\text{ and we compare with}\\ \widehat{B^*CD} &= \widehat{B^*CB} - \widehat{DCB} \ ,\qquad\text{ so let us compute the last angle first}\\ \widehat{DCB} &=\pi - \widehat{DMB} \\ &=\pi - \widehat{DMc} - \widehat{CMB} \\ &=\pi - \frac 12 \hat B - (\pi-\hat A) \\ &= \hat A - \frac 12 \hat B\ ,\qquad\text{so} \\ \widehat{B^*CD} &= \widehat{B^*CB} - \widehat{DCB} \\ &= \left( \hat C +\frac 12\hat B \right) - \left( \hat A - \frac 12 \hat B\right) \\ &=\hat B+\hat C -\hat A \\ &=\pi -2\hat A \\ &=2\widehat{NCD}\ . \end{aligned} $$
  • So $CN$ is the angle bisector of $\angle DCB^*$, which means we have the measure of $\overset \frown {NB^*}$ by considering (double of the) inscribed angles: $$ \widehat{NCB^*} = \widehat{DCN} = \frac\pi 2-\hat A = \widehat{ABB_2} = \widehat{ACC_2} \ , $$ so $$ \overset \frown{B^*N}= \overset \frown{B_2A}= \overset \frown{AC_2}\ , $$ $$ B^*N= B_2A= AC_2\ , $$ and this gives a construction of $N$, which is independent of $D$.

  • Observation: $B^*$ is the half point on the arc from $A$ to $C$, so $PB^*$ is the side bisector of $AC$. Then $$ \widehat{NPB^*} = \operatorname{measure}(\overset\frown{NB^*}) = \frac 12 \operatorname{measure}(\overset\frown{C_2AB_2}) \widehat{C_2BB_2} \ , $$ and because of $BB_2\|PB^*$ we get $PN\| BC_2$.

  • Note that from $ \overset\frown{C_2N} =\overset\frown{AB^*} =\overset\frown{B^*C} $, we get $NB^*\|CC2\|PQ$, where $PQ$ is the side bisector of $AB$, where $Q$ is the center of thew circle $(ABD)$, chosen by my hand on the side bisector for its reference, since we finally have to add it into the picture.

  • We have now everything but (details on) $Q$. Time to attack the point $Q$. (Try to stop here and find the own proof starting with the above information.) I will take a break here and insert some details (which are not important for the rest of the proof), just to not divulgate the idea in the very next line.


  • This is the break, the intermezzo. Observations without proof. The angle in $M$ between $CM$ and the prolongation of $AM$ is equal to $\angle AMC_1=\hat B$, so i also filled in an unneeded complication, the bisector of this angle. Then we have in $M$ three angles equal to $\frac12 \hat B$, so it is natural to draw all three of them. For this, consider the reflection $D'$ of $D$ w.r.t. $MC$. Then the three equal angles are between the prolongation of $AN$, then $MD'$, $MC$, $MD$. And the point $D'$ is on the side bisector $PB^*$!


  • Let us finally show $QP=NB^*$ by considering the projections of these segments, placed on parallel lines, onto the bisector $BB^*$. The projections of $Q,P$ on it are respectively the midpoints of $BD$, and respectively $BB^*$, so the length of the $PQ$-projected segment is $\frac 12(B^*B-DB)=\frac 12B^*D$, which is also the projection of $NB*$. From $QP$ parallel and equal $NB^*$ we get $PQNB^*$ parallelogram, so $QN=PB^*=PN$.

$\square$

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