## Wednesday, August 17, 2016

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1247. 1. Problem 1247
Let 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

2. Problem 1247 solution 2
Let 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 .Τherefore DM=MH.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

3. https://goo.gl/photos/k22CjrrhiiDy6UYo9

Note 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