Prove angle AGN = angle PGM'

Question

Given $\Delta ABC$ with its altitudes $AD,BE,CF$ and orthocenter $H$. Let $M,N,P$ the midpoint of $BC,AH,EF$. Let $G$ be the foot of perpendicular from $A$ to $EF$. Let $M'$ be the image of $M$ by reflection wrt $D$-midline of $\Delta DEF$. Prove $\angle AGN= \angle PGM'$

I thing angle chasing may solve the problem. But I kept chasing arround the conclusion. Also, the reflection wrt to the midline hypothesis is hard to approach. The only idea I have now is to prove $(NM'G)$ tangent to $EF$, but this is hard too.

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

1

Proof

It's well-known that $D,M,E,N,F~$lie on a circle, which is the nine-point circle of $\triangle ABC,$ and $A,E,H,F~$lie on another circle, whose center is $N$.

We may readily obtain that $M,P,M',N$ are collinear. From $D$ draw a line parallel to $EF$, which intersects $NM,AG$ at $Q,R$ respectively. Thus, $\triangle M'PG \cong \triangle MQR$. Let $AG$ intersect $BC$ at $T$, and $AD$ intersect $EF$ at $S.$ Obviously, $\triangle ASG \sim \triangle DTR.$

Notice that $\triangle FSN \sim \triangle DFN$. Thus$$\frac{AN}{NS}=\frac{NF}{NS}=\frac{ND}{NF}= \frac{ND}{AN}=\frac{DM}{MT}.$$

Hence, $\triangle ANG \sim \triangle DMR,$ which implies that $\angle PGM'=\angle DRM=\angle AGN.$

edited Jun 12 '20 at 10:38

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answered Aug 06 '18 at 12:18

mengdie1982

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What inspire you to have this solution? It's nice but I don't understand much the idea behind the solution. – RopuToran Aug 06 '18 at 16:07
any question is welcom！ – mengdie1982 Aug 06 '18 at 16:23
OK, I get it. Thank you. I suggest a small note: we have $(AHSD)=-1$, so $NA^2=NS.ND$ (we don't need to use a lot of similarity triangles) – RopuToran Aug 06 '18 at 17:22
sure,you can do that. but I want to give a solution as basic as possible. – mengdie1982 Aug 06 '18 at 17:45

Prove angle AGN = angle PGM'

1 Answers1

Proof