Geometry Parallel Proof Problem

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Geometry Parallel Proof Problem

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#1 h May 9, 2025, 3:33 AM • 1 Y

Y by Rounak_iitr

Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below

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#2 h May 9, 2025, 4:03 AM

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note that $D'$ is midpoint of arc $BC$
It is easy to see the line $DD'$ is perpendicular bisector of segment $EF$
tangent at $A$ to $(ABC)$ intersect $(ADM)$ and $BC$ at $X,Y$
known that $\angle YAD = \angle YDA$ implies that $DA \parallel MX$ .
will prove that $M,X,N$ collinear
Point $Z$ on $(ADM)$ such that $\angle BAZ = \angle MAC$ . so line $DD'$ is perpendicular bisector of segment $MZ$ .
$N' = MX \cap EF$
$-1=(AY,AZ;AB,AC)=(AX,AZ;AE,AF) = (X,Z;E,F) = (MX,MZ;ME,MF) = (N', \infty_{EF} ; E ,F)$ implies that $N'$ is midpoint of $EF$ .
So $MN \parallel AD$

This post has been edited 3 times. Last edited by ItzsleepyXD, May 9, 2025, 4:06 AM
Reason: typoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

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#3 h May 9, 2025, 4:37 AM

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Any other solutions..

This post has been edited 1 time. Last edited by CatalanThinker, May 9, 2025, 4:39 AM
Reason: typo

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#4 h May 9, 2025, 4:38 AM

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ItzsleepyXD wrote:

note that $D'$ is midpoint of arc $BC$
It is easy to see the line $DD'$ is perpendicular bisector of segment $EF$
tangent at $A$ to $(ABC)$ intersect $(ADM)$ and $BC$ at $X,Y$
known that $\angle YAD = \angle YDA$ implies that $DA \parallel MX$ .
will prove that $M,X,N$ collinear
Point $Z$ on $(ADM)$ such that $\angle BAZ = \angle MAC$ . so line $DD'$ is perpendicular bisector of segment $MZ$ .
$N' = MX \cap EF$
$-1=(AY,AZ;AB,AC)=(AX,AZ;AE,AF) = (X,Z;E,F) = (MX,MZ;ME,MF) = (N', \infty_{EF} ; E ,F)$ implies that $N'$ is midpoint of $EF$ .
So $MN \parallel AD$

Thanks, understood

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#5 h May 9, 2025, 4:42 AM

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Any other solutions?

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#6 h May 10, 2025, 2:52 PM

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$[asy] import graph; import geometry; size(9cm); defaultpen(fontsize(10pt)); pair A = (0,2); pair B = (-0.8,0); pair C = (3,0); pair M = (B+C)/2; pair in = unit(B-A)+unit(C-A)+A; pair ex = rotate(90,A)*in; pair D = extension(A,ex,B,C); path circ1 = circumcircle(A,B,C); path circ2 = circumcircle(A,D,M); pair F = intersectionpoints(A+3*(A-C)--A, circ2)[0]; pair E1 = intersectionpoints(B+3*(B-A)--A, circ2)[0]; pair N1 = (E1+F)/2; pair P = intersectionpoints(circ1,circ2)[1]; pair Q = extension(F,E1,D,C); path circ3 = circumcircle(Q,N1,M); path circ4 = circumcircle(Q,B,E1); draw(A--B--C--cycle, black); draw(A--F, black); draw(B--E1, black); draw(D--B, black); draw(E1--F, blue); draw(D--P, orange); draw(M--N1, royalblue); draw(A--D, royalblue); draw(circ1); draw(circ2, red); draw(circ3, deepgreen+dashed); draw(circ4, deepgreen+dashed); dot(A); dot(B); dot(C); dot(D); dot(M); dot(F); dot(E1); dot(N1); dot(P); dot(Q); label("$A$", A, N, black); label("$B$", B, S+1.25W, black); label("$C$", C, SE, black); label("$D$", D, W, black); label("$M$", M, NE, black); label("$E$", E1, S, black); label("$F$", F, N, black); label("$N$", N1, SW, black); label("$P$", P, SE, black); label("$Q$", Q, NW, black); [/asy]$
First, note that $D$ is the midarc $FE$ because $\angle{DAE}=\angle{DAF}$ .
Let $P\neq D$ be an intersection of the line $DN$ and $(ADM)$ , and $Q$ be an intersection of $\overline{FE}$ and $\overline{DM}$ .
Since $FPED$ is a harmonic quadrilateral with diameter $\overline{DP}$ , Picking ratio from $A$ to $\overline{DC}$ :
$(D,AP\cap DC;B,C)=-1.$ So, $\overline{AP}$ bisects $\angle{BAC}$ . Note that $\angle{DMP}=90^{\circ}$ , so $P\in (ABC)$ .
It is easy to see that $Q,N,M$ and $P$ are concyclic (since $\angle{QNP}=90^{\circ}=\angle{QMP}$ ).
Note that $A=BE\cap CF$ and $(AFE)$ meets $(ABC)$ again at $P$ .
Therefore $P$ is a spiral center sending $\overline{BC}\mapsto \overline{EF}$ . Now, we have $\triangle{PBC}\sim \triangle{PFE}$ .

Next, observe that $\angle{QEP}=\angle{PBC}$ . So, $Q,B,P$ and $E$ are concyclic.
Note that $Q=BM\cap NE$ , and $(QNM)$ meets $(QBE)$ again at $P$ .
So, $P$ is also a sprial center sending $\overline{BE}\mapsto \overline{MN}$ . Therefore, we have $\triangle{PNM}\sim \triangle{PEB}$ .
Since, $P$ sends $\overline{BC}\rightarrow\overline{EF}$ , $P$ also sends $\overline{BE}\rightarrow\overline{CF}$ . Which implies
$\triangle{FPC}\sim \triangle{EPB}\sim\triangle{NPM}.$ We must have $\angle{PNM}=\angle{PFC}=\angle{PFA}=\angle{PDA}.$ Therefore, $\overline{MN}\parallel\overline{AD}$ as desired.

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