Geometry question

Pure geometry question

CDCD is a chord of circle with centre OO.

ABAB diameter bisecting it.

OCOC drawn and its midpt KK

AKAK drawn intersecting circle at NN

Prove that NDND bisects BCBC

Note by Megh Parikh
5 years, 7 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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 DNBC=XDN\cap BC=X

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

Xuming Liang - 5 years, 7 months ago

Log in to reply

Thank you very much. very nice solution

Megh Parikh - 5 years, 7 months ago

Log in to reply

(y) Nice

Rohan Chandra - 5 years, 7 months ago

Log in to reply

Here goes my solution : Denote {M}=BCNK\{M\} = BC \cap NK then we have NCKMNCKM is concyclic (since KNM=OBC=KCM\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 NKM=NCM=NAB    KMAB\angle NKM = \angle NCM= \angle NAB \implies KM || AB . Now, consider the triangle OBCOBC with KMOBKM || OB and KK is the midpoint of OC, these lead to the fact that MM being the midpoint of BCBC as expected.

Viet Hoang - 5 years, 7 months ago

Log in to reply

Elegant solution

Megh Parikh - 5 years, 7 months ago

Log in to reply

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 DNBC=XDN\cap BC=X, we denote AKBC=MAK\cap BC=M.

If we want to prove XX is the midpoint of BCBC, we just need KXBOKX\parallel BO. Since OO is the midpoint of ABAB, we now just need to prove K(X,O;B,A)K(X,O;B,A) is a harmonic pencil     \iff so is K(X,C;B,M)K(X,C;B,M)     \iff so is N(X,C;B,M)N(X,C;B,M)     \iff ACBDACBD is a harmonic quadrilateral which is true since BDCA=DABCBD*CA=DA*BC. Q.E.D

Xuming Liang - 5 years, 7 months ago

Log 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....

Xuming Liang - 5 years, 7 months ago

Log in to reply

Thanks! :)

Happy Melodies - 5 years, 7 months ago

Log in to reply

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)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)K(X,C;B.M) to the ACBDACBD as a harmonic quad is correct, but then when you project the line XCBMXCBM, to the line OBAOBA, through point KK, you did not project XX. Cheers!

Yong See Foo - 5 years, 7 months ago

Log in to reply

Ok, I think I see the confusion here. Well, KXKX is considered to intersect ABAB at the point at infinity on ABAB, 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 - 5 years, 7 months ago

Log in to reply

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

Yong See Foo - 5 years, 7 months ago

Log in to reply

Bonus point to proving ANAN trisects BCBC, and another bonus point to proving DNDN trisects BOBO!

Yong See Foo - 5 years, 7 months ago

Log in to reply

Both are trivial using Menelaus' theorem.

Xuming Liang - 5 years, 7 months ago

Log in to reply

Yes menneeaus will prove it

Megh Parikh - 5 years, 7 months ago

Log in to reply

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

Yong See Foo - 5 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...