Thursday, July 30, 2009

Problem 330. Cyclic quadrilateral, Perpendicular diagonals, Area, Circumcenter

Proposed Problem
Click the figure below to see the complete problem 330 about Cyclic quadrilateral, Perpendicular diagonals, Area, Circumcenter.

 Problem 330. Cyclic quadrilateral, Perpendicular diagonals, Area, Circumcenter.
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Complete Problem 330
Level: High School, SAT Prep, College geometry

3 comments:

  1. Draw OM, ON perpendicular to AD, CD respectively. Given that AC is perpendicular to BD and that ABCD is cyclic, we can say, as a consequence of Brahmagupta’s Theorem, that OM = BC/2 and ON = AB/2.
    Now, Quad. AOCD = Tr. OAC + Tr. ODC = (1/2)(AD*OM+CD*ON)=(1/4)(AD*BC+CD*AB) ----------(1)
    By Ptolemy’s Theorem, AD*BC+CD*AB = BD*AC = BD (AE+EC) = BD*AE + BD*CE = 2(Tr.ABD+Tr.DCB)
    Or AD*BC+ CD*AB = 2(Tr.ABD+Tr.DCB)=2(Quad ABCD) --------(2)
    Combining equations (1) & (2), we’ve: Quad. AOCD = (1/4)( 2*Quad ABCD) = (1/2)*(Quad. ABCD)
    Now, Quad. ABCO = Quad ABCD – Quad AOCD =Quad. ABCD – (1/2)*(Qud. ABCD) =(1/2)*(Quad. ABCD)
    Or Quad AOCD = Quad. ABCO.
    Vihaan: vihaanup@gmail.com

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  2. Thank you, you have been brilliant keep posting.

    Robert

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  3. Trianlges BCO and ADO have the same area becauss angles BOC and AOD are supplementary.
    So are triangles ABO and CDO.
    That ends a proof.

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