If in two `DeltaABC` and `DeltaPQR`, `(AB)/(QR)=(BC)/(PR)=(CA)/(PQ)`, then

In DeltaABC and DeltaPQR , if (AB)/(QR)=(BC)/(PR)=(CA)/(PQ) , then

If in two triangles ABC and PQR, (AB)/(QR)=(BC)/(PR)=(CA)/(PQ) , then

If in triangleABC and trianglePQR we have (AB)/(QR)=(BC)/(PR)=(CA)/(PQ) then

DeltaABC~DeltaPQR,(AB)/(PQ)=(7)/(5) then ……………

In DeltaABC and DeltaPQR , in a one to one correspondence. (AB)/(QR) = (BC)/(PR) = (CA)/(PQ) , then

In DeltaABC and DeltaPQR , in a one to one correspondence. (AB)/(QR) = (BC)/(PR) = (CA)/(PQ) , then

In DeltaABC and DeltaPQR in a one-to-one correspndence (AB)/(QR)=(BC)/(PR)=(CA)/(PQ) then

In DeltaABC and DeltaPQR , in a one-to-one correspondance (AB)/(QR)=(BC)/(PR)=(CA)/(PQ) then……

If in DeltaABCandDeltaPQR for some one- one correpondence if (AB)/(QR)=(BC)/(PR)=(CA)/(PQ) then

Given that DeltaABC~DeltaPQR , CM and RN are respectively the medians of DeltaABC and DeltaPQR . Prove that (CM)/(RN)=(AB)/(PQ)