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

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

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## Saturday, January 16, 2016

### Geometry Problem 1180: Quadrilateral, 120 Degrees, Diagonals, Perpendicular

Labels:
120,
angle,
diagonal,
perpendicular,
quadrilateral

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Let AB, DC meet at H so that E is the incentre of Tr. AHD and < BGE = < DGE = 30.

ReplyDeleteAlso easily < ACD = 60 + A/2 = < DBG ( A + 60 - A/2) making BECH concylic

Hence BE = CE = 2FE since Tr . ECF is 30-60-90.

Further Tr. s GED is congruent to Tr. CED case ASA making EG = EC

I am not sure from "E is the incentre of Tr. AHD" , you get " < BGE = < DGE = 30.

Deleteplease explain .

Peter

Just a typo replace < BGE = < DGE = 30 with < BHE = < DHE = 30

DeleteI could not complete the last line

ReplyDeleteHence EG = 2FE

Sumith Peiris

Moratuwa

Sri Lanka

http://s1.postimg.org/kgkafchkf/pro_1180.png

ReplyDeleteBA meet CD at M

Since ∠(A)+ ∠(D)= 120 =>∠(M)=60

Since EA and ED bisect ∠(A) and ∠(D) => E is in center if trị. MAD and ME bisect ∠((M) and ∠( AED)=120

MBEC is cyclic and chord BE=chord EC => trị BED is isosceles

Triangle EFC is 30-60-90 triangle => EC= 2. EF

Triangles ECD and EDG are congruence…. ( case ASA) => EG=EC=2. EF

To Sumith & Peter

ReplyDeleteWhy Tr CED congrc to EGD

ASA. ED common. <GDE = <CDE and < GED = < CED = 60 which I'm sure u can figure out why.

ReplyDeleteOK to Sumith

ReplyDeleteFrom congr in tr BGC, GF median and BM median (M on GC)

=>EG = 2FE

Interestingly Tr. BGC is equilateral from which the result EG = 2 FE is obvious.

ReplyDelete