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

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

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## Wednesday, August 17, 2016

### Geometry Problem 1247: Triangle, Circumcircle, Orthocenter, Altitude, Perpendicular, Midpoint

Labels:
altitude,
circle,
circumcircle,
midpoint,
orthocenter,
perpendicular,
triangle

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Problem 1247

ReplyDeleteLet DE ,AH intersects the circumcircle of the triangle ABC in points K,L respectively.

Let N symmetry as D to BC then DE=EN ( H,L are symmetrical to BC ).So DHLN is

isosceles trapezoid. Then <HND=<LDN=<LDK. But AL//DK so <LDK=<AKD or <HND=<AKD

or AK//HN. But <DFC=90=<DEC so DFCE is cyclic , then <FED=<FCD=<ACD=<AKD so

FE//AK or FE//HN.Therefore DM=MH.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Problem 1247 solution 2

ReplyDeleteLet CH intersect the circumcircle of the triangle ABC in point K and EF in N.If EF intersect

the AB at G then DG is perpendicular AB (line EFG is Simson). K,H are symmetrical to AB then <GKH=<GHK.Is AGFD cyclic so <DGF=<DAF or <DGN=<DAC=<DKC=<DKN then DGKN is cyclic (DG//KN ). So DGKN is isosceles trapezoid.Then <DNK=<GKN=<GHK ie GH//DN.

But DG//HN so DGHN is parallelogram .Ī¤herefore DM=MH.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

https://goo.gl/photos/k22CjrrhiiDy6UYo9

ReplyDeleteNote that EF is the Simson line of point D.

From H draw a line parallel to EF

This line is Steiner line of point D .

Steiner line is the image of Simson line in the homothetic transformation with factor= 2 ( property of Steiner line)

So DF= ½ DL ,

DK= ½ DM

So DM= ½ DH = > M is the midpoint of DH