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