In the figure, AB is parallel to PRQ, T and P are the midpoints of CQ and CA respectively and ART is a straight line. Prove that TR : RA = 1 : 2.

Solution

Point Q is the midpoint of BC (midpoint theorem)

QB : QC QC : TQ

1 : 1 2 : 1

x2 x1

2 : 2 2 : 1

QB : TQ

2 : 1

∠TRQ = ∠TAB (corresponding RQ//AB)

TQ/QB = TR/RA

1/2 = TR/RA

Therefore TR : RA = 1 : 2

∠TQR = ∠TBA (corresponding ∠s, RQ//AB)

Therefore triangle TRQ is similar to triangle TAB