If `AB=QR`, `BC=PR` and `CA=PQ` then

In Delta DEF and Delta PQR , if PQ =DE, EF=PR and FD =QR, then

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

If in triangle ABC and triangle PQR for some one-one correspondence if (AB)/(QR)=(BC)/(PR)=(CA)/(PQ) then: a) triangle PQR~triangle ABC b) triangle PQR~triangle CAB c) triangle CBA~triangle PQR d) triangle BCA~triangle PQR

In quad PQR,AB=PQ,BC=QR and CB and RQ and PQR,AB=PQ,BC=QR and CB and RQ are extended to X and Y respectively and /_ABX=/_PQY .Prove that ABC~=PQR

Triangles ABC and PQR are both isosceles with AB=AC and PO=PR respectively.If also,AB=PQ and BC=QR, are the two triangles congruent? Which condition do you use? If /_B=50^(@), what is the measure of /_R?

If in triangleABC and trianglePQR we have (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