The midpoint of a side

by liberator, Sep 7, 2014, 3:39 PM

Problem: Let $ABCD$ be a cyclic quadrilateral. Denote $P=AD\cap BC, Q=AC\cap BD,$ $R=PQ\cap AB$.Let $M\neq R$ be the intersection of the circumcircle of $\triangle CDR$ and $AB$. Prove that $M$ is the midpoint of the side $AB$.

[b]My solution[/b]
This post has been edited 3 times. Last edited by liberator, Sep 7, 2014, 6:38 PM

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why AK,BJ,PM concur?

by bcp123, Sep 19, 2014, 6:45 PM

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They concur by cyclocevian conjugacy.

Theorem Let $P$ be defined by the cevian rays $AL, BM, CN$. Let circle $(LMN)$ intersect $BC,CA,AB$ again at $L', M', N'$ respectively. Then $AL', BM', CN'$ concur.

[asy]
/* Free script by liberator, 21 September 2014 */
unitsize(3cm);
pointpen=black;
pen dsp = rgb(0.4,0.6,0.8);
pen psp = rgb(0.8,0,0.8);
pen rcp = rgb(0.9,0,0);
/* Initialize objects */
pair A = dir(200);
pair B = dir(340);
pair C = dir(60);
pair D = dir(100);
pair P = IntersectionPoint(Line(B,C,2014,2014), Line(A,D,2014,2014));
pair Q = IntersectionPoint(A--C, B--D);
pair R = IntersectionPoint(Line(P,Q,2014,2014), A--B);
pair M = midpoint(A--B);
pair J = IntersectionPoint(circumcircle(C,D,R), D--A, 1);
pair K = IntersectionPoint(circumcircle(C,D,R), B--C, 1);
draw(A--B--P--cycle, dsp+linewidth(1));
draw(P--C, dsp+linewidth(1));
draw(P--D, dsp+linewidth(1));
draw(P--R, dsp);
draw(A--C, dsp);
draw(B--D, dsp);
draw(A--K, psp);
draw(B--J, psp);
draw(P--M, psp);
draw(circumcircle(C,D,R), rcp);
/* Place dots on and label each point */
Drawing("A", A, dir(A));
Drawing("B", B, dir(B));
Drawing("L", C, dir(C));
Drawing("M", D, dir(110));
Drawing("C", P, dir(P));
Drawing("P", Q, dir(0));
Drawing("N", R, dir(-70));
Drawing("M'", J, dir(180));
Drawing("L'", K, dir(0));
Drawing("N'", M, dir(-110));
Drawing("P'", intersectionpoint(A--K, B--J), 1.5*dir(120));[/asy]
Proof By intersecting chords,

\[\frac{AN \cdot AN'}{MA \cdot MA'} \cdot \frac{BL \cdot BL'}{BN \cdot BN'} \cdot \frac{CM \cdot CM'}{CL \cdot CL'} = 1.\]
But $\frac{AN}{NB} \cdot \frac{BL}{LC} \cdot \frac{CM}{MA} = 1$, hence $\frac{AN'}{N'B} \cdot \frac{BL'}{L'C} \cdot \frac{CM'}{M'A} = 1$, so $AL', BM', CN'$ concur. $\Box$
This post has been edited 1 time. Last edited by liberator, Sep 27, 2014, 8:36 PM

by liberator, Sep 21, 2014, 6:24 PM

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