Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 782.
Sunday, July 15, 2012
Problem 782: Triangle, Orthocenter, Circumcenter, Circumcircle, Diameter, Altitude, Parallelogram
Labels:
altitude,
circumcenter,
circumcircle,
diameter,
orthocenter,
parallelogram,
triangle
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AD diameter => ABD=90 degrees
ReplyDelete=> BD//HC
AD diameter => ACD=90 degrees
=> DC//BH
Hence, BDCH is parallelogram.
let the altitudes from A,B,C touches the opposite sides at P,Q,R.
ReplyDeleteAngle PHC = Angle ABC = Angle ADC
Angle BHP = Angle ACB = Angle BDA
So Angle BHC = Angle BHP + Angle BHP = Angle ADC + Angle BDA = Angle BDC
Angle ABQ = Angle ACR
Hence, Angle HBD = 90 - Angle ABQ = 90 - Angle ACR = Angle HCD
So we have proved two same pairs of opposite angles.
Q.E.D.
(Alternative way by Vector)
ReplyDeletelet vector OA,OB,OC be a,b,c respectively.
Then vector OH = a+b+c, OD = -a
vector BD = OD - OB = -a-b
vector HC = OC - OH = c - (a+b+c) = -a-b
Hence vector BD = vector HC
A parallelogram is formed
q.e.d.
Let M be the midpoint of AB
ReplyDeleteCH // OM // BD and
CH = 2 OM = BD
Follows BDCH is a parallelogram