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

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

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## Sunday, November 11, 2018

### Geometry Problem 1399: Triangle, Circumcircle, Perpendiculars, Angles, Congruence, Cyclic Quadrilateral

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Since CEFG is concyclic,

ReplyDelete< ECF = < EGF = x say

Now since AGEC is concyclic,

< EGH = < EAH = < DCB

But < EGH = alpha + x and

< DCB = alpha’ + x and so

alpha = alpha’

Sumith Peiris

Moratuwa

Sri Lanka

Solution 2

ReplyDeleteSince AGEH & ABDC are both concyclic

< HEG = < BDC

So in Tr.s BDC & HEG,

< HEG = < BDC

< EHG = < DBC since each angle = < CAD

Hence < EGH = < DCB

But < EGF = < ECF

So alpha = alpha’

Sumith Peiris

Moratuwa

Sri Lanka

Ang BAD = HGF + FGE ( AHEG )

ReplyDeleteAng BAD = BCE + ECD ( BD )

Ang FGE = FCE ( FECG )

=> HGF = ECD

Unghiul FEG = unghiul FCG (laturi perpendiculare).

ReplyDeleteDe aici FECG circumscriptibil

=> unghiul FGE = FCE (1).

Unghiul AHE = unghiul AGE =90

=> AHEG circumscriptibil

Unghiul HAE = unghiul HGE . (2)

Dar unghiul BAD = unghiul BCD (3)

Din (2) si (3) => unghi BCD= unghi HGE.(4)

Din (1) si (4) => α= α^,

Florin Popa, Comanesti, Romania

Is EFGC=cyclic=>α=<DCE=<FGH=α'.

ReplyDelete