Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 777.
http://img839.imageshack.us/img839/7122/problem777.pngDraw lines per attached sketchWe have LC//BH and LB//CE So BLCH is a parallelogram and BH=LC=AL. cos(ALC)= 2.R.Cos(ABC)
HB = BD / sin(BHD)= AB*cos(ABC) / sin(ACB)...CDHF concyclic= 2R*cos(ABC)...Sine rule
AC/sin(ABC) = 2R (R is the circumradius)The statement is then reduced to prove : AC/sin(ABC) = BH/cos(ABC) <=> AC = BH*tanABCwhich is the properties of the altitudeq.e.d.
Let M be the midpoint of AC.Median BM intersects Euler line OH at centroid G.∆BGH ~ ∆MOG since OM ∥ BHSo BH/OM = BG/GM = 2 andBH = 2OM = 2.OA.cos∠AOM = 2R cos∠ABC [since∠AOM=(1/2)∠AOC=(1/2)(2∠ABC)=∠ABC]
Since < BHD = < C, BD = BH sin <C ...,(1)Also AB = 2R sin <C ....(2)And BD = AB cos <B ...(3)Eliminating AB and BD between these 3 equationsBH = 2R cos < BSumith PeirisMoratuwaSri Lanka
Alternatively let AO meet the circle at X then prove that BHCX is a parellelogram (refer my proof of Problem 778)Then BH = CX = 2R cos <B since < AXC = < B