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.
Wednesday, August 17, 2016
Geometry Problem 1247: Triangle, Circumcircle, Orthocenter, Altitude, Perpendicular, Midpoint
Subscribe to:
Post Comments (Atom)
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