The sides AB,BC,CA of triangle ABC touch a circle with centre o and radius r at P,Q,R respectively.Prove that

(1) AB+CQ= AC+BQ

(2) Area (triangle ABC) = 1/2 (perimeter of triangle ABC) x r (radius)

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  Join AO , BO , OC

PO = OQ = OR = r

TANGENTS FROM AN EXTERNAL POINTS R EQUAL

therefore

AP = RA  -----1

PB = BQ

RC = QC ----2

1 + 2

AP + RC = RA + QC

Adding BQ onboth sides

AP + RC + BQ = RA + QC + BQ

Since BQ = PB      QC = RC    

AP + CQ + PB = RA + RC + BQ

AB + CQ =AC + BQ
 
 
2
Area (tri . ABC) = Area (tri .AOB) + Area (tri .BOC) + Area (tri . COA)
                           = 1/2  AB x r  + 1/2 BC x r  + 1/2 CA x r
                           = 1/2 r (AB + BC + CA)
                           -= 1/2 r ( perimeter of triangle ABC )

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