In triangle PQR, a line XY parallel to QR cuts PQ at X and PR at Y in which PX/XQ=PY/YR=1/2.Then
a)XY=QR
b)XY=1/2QR
Consider ΔPQR and ΔPXY
∠PQR=∠PXY ( corresponding angles and XY is parallel to QR)
∠QPR= ∠XPY (common)
Therefore ΔPQR∼ΔPXY
This implies