Geometry Problem. Post your solution in the comments box below.

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

Click the figure below to view the complete problem 995.

## Friday, March 21, 2014

### Geometry Problem 995: Quadrilateral, Perpendicular Diagonals, Perpendicular Bisector, Parallel

Labels:
diagonal,
parallel,
perpendicular,
perpendicular bisector,
quadrilateral

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Since ∆AEC and ∆BFC are isosceles triangles

ReplyDeleteSo ∠BFD=180- 2. ∠ODA

And ∠CEA=180-2. ∠CAE

And ∠BFD+∠CEA= 360-2.( ∠ODA+∠CAE)= 180

So BF//CE

Problem 995

ReplyDeleteLet <EAC=<ECA=x,<FBD=<FDBX+Y=y.But in triangle AOD apply <DAO+<ADO=90 or x+y=90. If BF intersects the AC at K then <BKO=90-<KBO=90-y=x=<AKF=<ACF. So

BF//CE.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

Let < NFB = p = < NFD = < CAE = < ACE

ReplyDeleteSo < CED = 2p = < BFP and so BF // CE

Sumith Peiris

Moratuwa

Sri Lanka

Let m(MEA)=@(Read as Alpha)=>m(NFE)=90-@

ReplyDeleteAME congruent to CME (SAS AM=MC,m(AME)=m(EMC)=90)

=> m(MEC)=m(MEA)=@ and m(AEC)=2@ -------(1)

Simiarly FNB congruent to FND

=> m(BFN)=m(NFD)=90-@=>m(AFB)=2@--------(2)

From (1)&(2) BF//CE