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If O is the circumcentre and O' the orthocenter of DeltaABC prove that (i) SA+SB+SC=3SG, where S is any point in the plane of DeltaABC. (ii) OA+OB+OC=OO' Where, AP is diameter of the circumcircle. |
Answer» Solution :Let G be the centroid of `DeltaABC`, first we shall show that CIRCUMCENTRE O, ORTHOCENTER O' and centroid G are collinear and O'G=2OG. Let AL and BM be perpendiculars on the sides BC and CA, respectively. Let AD be the median and OD be the perpendicular from O on side BC. iff R is the circumradius of circumcircle of `DeltaABC`, then OB=OC=R. In `DeltaOBD,` we have OD=R cosA. . . (i) In `DeltaABM,AM=AB cos A=c cosA`. . . (ii) FORM `DeltaAO'M,AO'=AMsec(90^(@)-C)` =c cos A cosec C `=(c)/(SINC)*cosA=2RcosA ""(because(a)/(sinA)=(b)/(sinB)=(c)/(sinC)=2R)` `AO'=2(OD)`. . . (iii) Now, `DeltaAGO' and DeltaOGD` are similar. `therefore(OG)/(O*G)=(GD)/(GA)=(OD)/(AO')=(1)/(2)`[using Eq. (iii)] `implies 2OG=O'G` (i) We have, `SA+SB+SC=SA+(SB+SC)` `=SA+2SD`(`because` D is the mid-point of BC) `=(1+2)+SG=8SG`. (ii) On REPLACING S by O in Eq. (i), we get `OA+OB+OC=3OG` `=2OG+OG=GO'+OG` `=OG+GO'+O O'`. |
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