Parallelogram PQRS has PQ = RS = 8 cm and diagonal QS = 10 cm. Point F is on RS, exactly 5 cm from S. Let T be the intersection of PF and QS. Draw a diagram and find the lengths of TS and TQ.
Parallelogram PQRS has PQ = RS = 8 cm and diagonal QS = 10 cm.
If you sketch the parallelogram as described there will be triangles FTS and PTQ within. FS =5 and PQ= 8. Let TS =x and TQ = 10 - x, since SQ=10. There are 3 pairs of congruent angles.
Angle FTS is congruent to angle PTQ by vertical angles.
Angle TFS is congruent to angle TPQ by alternate interior angles.
Angle TSF is congruent to angle TQP by alternate interior angles.
Triangles FTS and PTQ are similar triangles, so corresponding side lengths are in proportion.
Segment SF corresponds to segment QP and segment TS corresponds to TQ.
Set up the proportion and 5/8 = x/(10 - x). Solve for x and 10 - x to find TS and TQ, respectively.