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Here goes my solution :
Denote $\{M\} = BC \cap NK$ then we have $NCKM$ is concyclic (since $\angle KNM = \angle OBC = \angle KCM$ : I skipped a few details here, you can verify them if you will). Using the angle chasing technique, one has $\angle NKM = \angle NCM= \angle NAB \implies KM || AB$. Now, consider the triangle $OBC$ with $KM || OB$ and $K$ is the midpoint of OC, these lead to the fact that $M$ being the midpoint of $BC$ as expected.

I will provide another solution just for fun. This one I'll use something called Harmonic division, for those that are unfamiliar just google, there's a lot of papers written on this wonderful topic. Here is the 2nd proof:

Keeping $DN\cap BC=X$, we denote $AK\cap BC=M$.

If we want to prove $X$ is the midpoint of $BC$, we just need $KX\parallel BO$. Since $O$ is the midpoint of $AB$, we now just need to prove $K(X,O;B,A)$ is a harmonic pencil $\iff$ so is $K(X,C;B,M)$$\iff$ so is $N(X,C;B,M)$$\iff$$ACBD$ is a harmonic quadrilateral which is true since $BD*CA=DA*BC$. Q.E.D

These are cross ratios right? To my knowledge, I think the way you do it is wrong? Because you stated $K(X,O;B,A)$ is a harmonic pencil, but they are not concyclic or collinear, can you do that? The part from $K(X,C;B.M)$ to the $ACBD$ as a harmonic quad is correct, but then when you project the line $XCBM$, to the line $OBA$, through point $K$, you did not project $X$. Cheers!

Ok, I think I see the confusion here. Well, $KX$ is considered to intersect $AB$ at the point at infinity on $AB$, which is something new introduced in projective geometry. Despite that though, it can be checked that the cross-ratio(trig ratio) isn't affect when you have parallel lines.

@Xuming Liang
–
Alright I remember now because given two points, if the midpoint divides the segment internally, the point at infinity on that line divides it externally with the same ratio, 1. So in $K(X,O;B,A)$, you actually meant $K(KX\cap AB,O;B,A)$, where $KX\cap AB$ is actually point at infinity, and therefore the two lines are parallel. Thanks for that!

I won't explain the notations here, since it's pretty "universal" in projective geometry. Again, I encourage those that have never heard of Harmonic divisions to read about it.

Although the words "projective geometry" might sound intimidating to some of you, in terms of just harmonic division, I think we can keep everything Euclidean by just knowing the fact that cross-ratios are possible due to the ratios of trig angles etc. I know these make no sense but you will know what I mean once you read about it....

This problem is based around this property for similar triangles: Corresponding elements in similar triangles are also "similar". This will be disguised in the following proof(more like a sketch actually):

Denote $DN\cap BC=X$

First note that $\triangle COA\sim \triangle CBD$. By properties of circles: $\angle CDX=\angle CAK,\angle BDX=\angle OAK$, using this we can prove $\triangle CDX\sim \triangle CAK, \triangle XDB\sim \triangle KAO$, which gives: $\frac {CX}{CK}=\frac {CD}{CA}=\frac {BD}{OA}=\frac {BX}{OK}$$\Rightarrow \frac {CX}{BX}=\frac {CK}{OK}=1$$\Rightarrow CX=BX$. Q.E.D

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## Comments

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TopNewestBonus point to proving $AN$ trisects $BC$, and another bonus point to proving $DN$ trisects $BO$!

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Both are trivial using Menelaus' theorem.

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Well using your notation I just play around with the harmonic conjugates $(X,C)$ and $(B,M)$, so it's pretty much the same thing with Menelaus'.

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Yes menneeaus will prove it

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Here goes my solution : Denote $\{M\} = BC \cap NK$ then we have $NCKM$ is concyclic (since $\angle KNM = \angle OBC = \angle KCM$ : I skipped a few details here, you can verify them if you will). Using the angle chasing technique, one has $\angle NKM = \angle NCM= \angle NAB \implies KM || AB$. Now, consider the triangle $OBC$ with $KM || OB$ and $K$ is the midpoint of OC, these lead to the fact that $M$ being the midpoint of $BC$ as expected.

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Elegant solution

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I will provide another solution just for fun. This one I'll use something called Harmonic division, for those that are unfamiliar just google, there's a lot of papers written on this wonderful topic. Here is the 2nd proof:

Keeping $DN\cap BC=X$, we denote $AK\cap BC=M$.

If we want to prove $X$ is the midpoint of $BC$, we just need $KX\parallel BO$. Since $O$ is the midpoint of $AB$, we now just need to prove $K(X,O;B,A)$ is a harmonic pencil $\iff$ so is $K(X,C;B,M)$ $\iff$ so is $N(X,C;B,M)$ $\iff$ $ACBD$ is a harmonic quadrilateral which is true since $BD*CA=DA*BC$. Q.E.D

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These are cross ratios right? To my knowledge, I think the way you do it is wrong? Because you stated $K(X,O;B,A)$ is a harmonic pencil, but they are not concyclic or collinear, can you do that? The part from $K(X,C;B.M)$ to the $ACBD$ as a harmonic quad is correct, but then when you project the line $XCBM$, to the line $OBA$, through point $K$, you did not project $X$. Cheers!

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Ok, I think I see the confusion here. Well, $KX$ is considered to intersect $AB$ at the point at infinity on $AB$, which is something new introduced in projective geometry. Despite that though, it can be checked that the cross-ratio(trig ratio) isn't affect when you have parallel lines.

Log in to reply

$K(X,O;B,A)$, you actually meant $K(KX\cap AB,O;B,A)$, where $KX\cap AB$ is actually point at infinity, and therefore the two lines are parallel. Thanks for that!

Alright I remember now because given two points, if the midpoint divides the segment internally, the point at infinity on that line divides it externally with the same ratio, 1. So inLog in to reply

I won't explain the notations here, since it's pretty "universal" in projective geometry. Again, I encourage those that have never heard of Harmonic divisions to read about it.

Although the words "projective geometry" might sound intimidating to some of you, in terms of just harmonic division, I think we can keep everything Euclidean by just knowing the fact that cross-ratios are possible due to the ratios of trig angles etc. I know these make no sense but you will know what I mean once you read about it....

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Thanks! :)

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This problem is based around this property for similar triangles: Corresponding elements in similar triangles are also "similar". This will be disguised in the following proof(more like a sketch actually):

Denote $DN\cap BC=X$

First note that $\triangle COA\sim \triangle CBD$. By properties of circles: $\angle CDX=\angle CAK,\angle BDX=\angle OAK$, using this we can prove $\triangle CDX\sim \triangle CAK, \triangle XDB\sim \triangle KAO$, which gives: $\frac {CX}{CK}=\frac {CD}{CA}=\frac {BD}{OA}=\frac {BX}{OK}$ $\Rightarrow \frac {CX}{BX}=\frac {CK}{OK}=1$ $\Rightarrow CX=BX$. Q.E.D

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Thank you very much. very nice solution

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(y) Nice

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