Proposed Problem
Click the figure below to see the complete problem 347 about Triangle, Altitude, Perpendicular, Circle, Concyclic points.
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Complete Problem 347
Level: High School, SAT Prep, College geometry
Monday, August 24, 2009
Problem 347. Triangle, Altitude, Perpendicular, Circle, Concyclic points.
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BF BC = BD squared = BE BA,
ReplyDeleteBF BC = BE BA
By the converse of power theorem, AEFC is cocycle.
Quad. BEDF is clearly concyclic; hence angle BEF = angle BDF = 90 - angle FDC = angle C. Hence AEFC is concyclic. QED.
ReplyDeleteVihaan
angle BED + angle BFD = 180 degrees because DE and DF are perpendiculars to AB and BC respectively. this implies that BEDF is a cyclic quadrilateral. join EF then angle DEF = angle DBF as they are in the same segment. but angle DBF = 90 - angle C as BDC is a right triangle. so angle DEF = 90 - angle C. this implies that angle AEF = 90 + 90 - angle C = 180 - angle C. so angle AEF + angle ACF = 180. so AEFC is a cyclic quadrilateral because the sum of a pair of opposite angles equals 180 degrees. therefore points A, E, F and C are concyclic.
ReplyDeleteQ. E. D.
Since EBFD is cyclic < BEF = < BDF which is in turn = to FCD.
ReplyDeleteHence AEFC is con cyclic
Sumith Peiris
Moratuwa
Sri Lanka
Let <BAD=x
ReplyDelete<EDA=90-x
<EDB=90-(90-x)=x
Since <BED=<BFD=90
<BED+<BFD=180 and hence BEDF is a cyclic quad
Join EF
<EFB=<EDB=x (< in same seg.)
So <EAD=<BAD=<EFB
So EACF is a cyclic quad (converse of < in alt seg.)