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

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

Click the figure below to view more details of problem 1167.

## Thursday, November 19, 2015

### Geometry Problem 1167: Triangle, Circle, Circumcircle, Perpendicular, 90 Degrees, Collinear Points

Labels:
90,
circle,
circumcircle,
collinear,
degree,
perpendicular,
triangle

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Since D is on circumcicle of ABC => GHN is Simson line from D

ReplyDeleteSo G,H,N are collinear

Since D is on circumcicle of AEF => GHM is Simson line from D

So G,H,M are collinear => G,H,M,N are collinear

< DCH = < DAG = < GHD and since DHMC is con cyclic G,H,N are collinear

ReplyDeleteFurther < GDM = < ADF and because GDME and ADFE are cyclic < ADG = < FDM and so < AHG = < MHF so G,H, M are collinear

Hence G,H, N, M are collinear

Sumith Peiris

Moratuwa

Sri Lanka

Sumith,

ReplyDeleteThere's seems to be a typo in your proof. Should it not be < DCB (not DCH) = < DAG = < GHD? As also D,H,N,C are concyclic not DHMC? Correct me if I'm wrong.

By the way, I need your assistance in planning a trip to Sri Lanka coming summer in April/May 2016. Would you be kind enough to send me a message at: ajitathle@gmail.com?

Ajit

Sorry Ajith I saw your post only now

ReplyDeleteBoth typos u pointed out are correct. Thank u

Welcome to Sri Lanka. I have already sent u a Test mail. Do contact me if u do not receive same