Proposed Problem

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

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Complete Problem 330

Level: High School, SAT Prep, College geometry

## Thursday, July 30, 2009

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

Labels:
area,
circumcenter,
cyclic quadrilateral,
diagonal,
perpendicular

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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.

ReplyDeleteNow, 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

Thank you, you have been brilliant keep posting.

ReplyDeleteRobert

Trianlges BCO and ADO have the same area becauss angles BOC and AOD are supplementary.

ReplyDeleteSo are triangles ABO and CDO.

That ends a proof.

Did it the same way.

DeleteThis proof though using trigonometry is the easiest