Challenging Geometry Puzzle: Problem 1551. Share your solution by posting it in the comment box provided.

Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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< FBE = < FAE = < CAD = < FGE = a (say)

ReplyDelete< JHE = < HEJ = < BDH = < BDA = b (say)

< FEB = < FAB = < FGB = u (say)

< HEC = < HJC = < HDC = v (say)

< ABG = < AFG = < AEG = t (say)

< JCD = < JHD = < JED = y (say)

a + b = 52.........(1) (Triangle BCE)

u + v + 52 + 46 = 180 (angles on Point E)

So u +v = 82 .......(2)

u + 2a + t = 180 - 52 = 128...(3) (Angles on Point G)

v + 2b + y = 180 - 52 = 128 ...(4) (Angles on Point J)

(3)+(4) ==> u + v + 2a + 2b + t +y = 256

So 82 + 104 + t + y = 256

t + y = 70

< GEJ = 180 - (a+t) - (b+y) = 180 - (a+b) - (t+y) = 180 - 52 - 70

< GEJ = 58

Sumith Peiris

Moratuwa

Sri Lanka

Because < BDC= < BAC;

ReplyDeleteSo, < HFE= < FHE

Same for <EGJ = EJG

So < FHE =90-46/2 = 67 degree

So < EDC = 67 degree

So < ECD =180 -52-67 = 61 degree

So < EJG =61 degree

So <GEJ = 2*(90-61)= 58 degree

Antonio - a new problem arises from this problem

ReplyDeleteProve that FHJG is a cyclic quadrilateral