## Sunday, July 15, 2012

### Problem 782: Triangle, Orthocenter, Circumcenter, Circumcircle, Diameter, Altitude, Parallelogram

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

Click the figure below to see the complete problem 782.

1. AD diameter => ABD=90 degrees
=> BD//HC

=> DC//BH

Hence, BDCH is parallelogram.

2. let the altitudes from A,B,C touches the opposite sides at P,Q,R.
Angle 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.

3. (Alternative way by Vector)
let 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.

4. Let M be the midpoint of AB
CH // OM // BD and
CH = 2 OM = BD
Follows BDCH is a parallelogram