1.

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`

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.


Discussion

No Comment Found