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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 782.
AD diameter => ABD=90 degrees=> BD//HCAD diameter => ACD=90 degrees=> DC//BHHence, BDCH is parallelogram.
let the altitudes from A,B,C touches the opposite sides at P,Q,R. Angle PHC = Angle ABC = Angle ADCAngle BHP = Angle ACB = Angle BDASo Angle BHC = Angle BHP + Angle BHP = Angle ADC + Angle BDA = Angle BDCAngle ABQ = Angle ACRHence, Angle HBD = 90 - Angle ABQ = 90 - Angle ACR = Angle HCDSo we have proved two same pairs of opposite angles.Q.E.D.
(Alternative way by Vector)let vector OA,OB,OC be a,b,c respectively.Then vector OH = a+b+c, OD = -avector BD = OD - OB = -a-bvector HC = OC - OH = c - (a+b+c) = -a-bHence vector BD = vector HC A parallelogram is formedq.e.d.
Let M be the midpoint of ABCH // OM // BD and CH = 2 OM = BDFollows BDCH is a parallelogram