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