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
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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
https://photos.app.goo.gl/eHb378Pg8qn43sM2A
ReplyDeleteLet M is the midpoint of BD
S(ABCO)=S(ABCM)= 1/2. AC.BM
S(ABCD)=1/2AC.BD= AC.BM= 2.S(ABCO)
so S(ABCO)= S(AOCD)