we have AB=b-a,AC=c-a=(1-r)a+rb-a=r(b-a)
BC=c-b=(1-r)a+rb-b=(r-1)b-(r-1)a =(r-1)(b-a)
PQ=v-u
PR=w-u=(1-r)u+rv-u=r(v-u)
QR=w-v=(1-r)u+rv-v=(1-r)u+(r-1)v=(r-1)(v-u)
Finally
so Triangles ABC and PQR are similar
Thanks and Regards
Shaik Aasif
askIITians Faculty