Proposed Problem
Click the figure below to see the complete problem 332 about Cyclic quadrilateral, Ratio of diagonals, Similarity.
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Complete Problem 332
Level: High School, SAT Prep, College geometry
Thursday, July 30, 2009
Problem 332. Cyclic quadrilateral, Ratio of diagonals, Similarity
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Triangles ABE & CDE are similar so AB/CD=AE/DE. Likewise, triangles BCE & ADE are similar so AD/BC=AE/BE. Multiplying the two equations, we've: (AB/CD) * (AD/BC) = AE^2/(BE*DE). But BE*DE=AE*CE, since quad. ABCD is cyclic. So, (AB/CD) * (AD/BC) = AE^2/(AE*CE) = AE/CE or
ReplyDeleteAE/CE = AB/BC * AD/CD. QED
(AB*AD)/(BC*CD)=(AB*AD*sinA)/(BC*CD*sinB)=(the area of ABD)/(the area of BCD)
ReplyDeleteAE/CE=(AE*BD*sinE)/(CE*BD*sinE)=(the area of ABD)/(the area of BCD)
therefore (AB*AD)/(BC*CD)=AE/CE
Albert,
ReplyDeleteFirst line:
AB*AD)/(BC*CD)=(AB*AD*sinA)/(BC*CD*sinC)=(the area of ABD)/(the area of BCD)
we know that the ratio between the areas of two triangles which are lies in between two paralles lines and on the different bases is equal to the ratio of their bases. so AE /CE = ar of Tr ABE / ar of Tr BCE = ar of Tr ADE / ar of Tr CED (by ratio and proportion) AE / CE = ar TrABE+ ar TrADE / ar TrBCE + ar TrCED = ar TrABD / ar TrBCD = AB.AD SinBAD / BC.CD SinBCD = AB.AD / BC.CD (since ABCD is cyclic quadrilateral so SinBAD = SinBCD) hence AE/CE = AB.AD / BC.CD vijay9290009015@yahoo.in
ReplyDeleteTriangle BAE ~ Triangle CDE
ReplyDeleteBA/AE=CD/DE
AE=BA*DE/CD-------(1)
Triangle BCE ~ Triangle ADE
BC/CE=AD/DE
CE=BC*DE/AD--------(2)
(1)/(2):
AE/CE=BA*DE*AD/BC*DE*CD=BA*AD/BC*CD