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

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

Details: Click on the figure below.

## Wednesday, March 1, 2017

### Geometry Problem 1320: Triangle, Incircle, Tangent, Chord, Circle, Parallel, Perpendicular, Collinearity

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Extend HF at G , G on AB. Join G to K, meet DE at P. From 1315 DP=NE =>PKN isoceles, => GKH isoceles

ReplyDeleteDear Antonio

ReplyDeleteCan this be proved directly without using Pr 1315?

Yes

Deletehttps://goo.gl/photos/NtvVeEVnYmjzHqrLA

ReplyDeleteLet HF cut DB at L; LK and FA cut DE at N’ and G

Per result of problem 1315 we have

DG=MN

GM=NE , GN=ME=DM

triangle FGM similar to FAK so GM= KA.(FM/FK)

Triangle LDN’ similar to LAK so DN’= KA.( LN’/LK)= KA.(FM/FK)

So GM= DN’=NE => MN=MN’

Due to symmetry of LN’K to HNK over axis of symmetry BK => H,N ,K are collinear

Angle FHN=HNE=MNK, so they are colinear.

ReplyDeletePlease explain the reason of " HNE=MNK" in your solution

ReplyDeletePeter Tran