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Prove m:n theoremin a `Delta ABC`, a point D is taken on side BC such that BD:DC is m:n.Then prove that(1) `(m+n)cottheta= mcotalpha-ncotbeta`(2) `(m+n)cottheta= ncotB-mcotC` |
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Answer» 1)`(BD)/(DC)=m/n` `/_ADB=180-theta` `/_ABD=180-(alpha+180-theta)` `=theta-alpha`. `In/_ABD` `(BD)/sinalpha=(AD)/(sin(theta-alpha))` `In/_ADC` `(DC)/sinbeta=(AD)/(sin(theta-beta))` `((BD)/sinalpha)/((DC)/sinbeta)=((AD)/(sin(theta+alpha)))/((AD)/(sin(theta+beta)` `(BDsinbeta)/(DCsinalpha)=(sin(theta+beta))/(sin(theta-alpha))` `(msinbeta)/(nsinalpha)=(sinthetacosbeta+costhetasinbeta)/(sinthetacosalpha-costhetasinalpha)` Divide by`sinthetasinbetasinalpha` `m(cotalpha-cottheta)=n(cotbeta+cottheta)` `mcotalpha-mcottheta=ncotbeta+ncottheta` `mcotalpha-ncotalpha=cottheta(m+n)` Hence proved. 2)`/_ABD=/_ABC=B` `/_ACD=c` `/_BAD=(theta-beta)` `msin(theta+c)sinbeta=nsincsin(theta-beta)` `msinbeta(sinthetacosc+costhetasinc)=nsinc(sinthetacosbeta-costhetasinbeta)` Divide by`sinthetasinbetasinc` `m(cotc+cottheta)=n(cotbeta-cottheta)` `mcotc+mcottheta=ncotbeta-ncottheta` `cottheta(m+n)=ncotbeta-mcotc` Hence Proved. |
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