Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 871.

## Friday, April 19, 2013

### Problem 871: Brahmagupta's Theorem, Cyclic Quadrilateral, Perpendicular Diagonals, Midpoint

Labels:
Brahmagupta,
cyclic quadrilateral,
diagonal,
midpoint,
perpendicular

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Angle MCE = Angle EDF = Angle MED - 90 = Angle MEC

ReplyDeleteSo triangle MCE is isosceles and so as triangle BEM for symmetry.

Hence BM = EM = CM

Q.E.D.

My solution http://www.docstoc.com/docs/153902983/Problem-871

ReplyDeleteFar more general:

ReplyDeleteIf ABCD is cyclic quadrilateral and its diagonals cuts in E, then orthocenter of tr ABE, circumcenter of tr DEC and E itself are colinear.

In this particular case, EF contains the orthocenter of tr AED so must contain the circumcenter of tr BEC, but this is M.

Done.

Problem 871

ReplyDeleteLet x=<BCA=<BDA=<AEF (perpendicular line) or <MEC=x.Then ME=MC.

But <EBC=90-x=<BEM.So BM=EM.Therefore BM=MC.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE