In Δ ABC, P, Q are points on AB and AC respectively and PQ || BC. Prove that the median AD bisects PQ.
                             

Given: ABC is a triangle, PQ || BC; AD is the median which cuts PQ at R.
To prove: AD bisects PQ at R.
Proof: In Δ ABD; PR || BD
  AP = AR  (BPT)
  PB    RD
In Δ ACD, RQ || DC
∴ AR = AQ (BPT)
  RD    QC
In ΔAPR and ΔABD,
∠APR = ∠ABD (corresponding angles.)
∠ARP = ∠ADB (corresponding angles.)
∴ Δ APR is similar to Δ ABD (AA similarity)
∴ AP = AR = PR (corresponding sides of similar triangles are proportional)----(i) 
  AB    AD    BD
Similarly Δ ARQ is similar to Δ ADC
\\  AQ = AR = RQ -----(ii)
   AC   AD    DC 
According to equation (i) and (ii),
 AR = PR = RQ
  AD   BD    DC
but BD = DC (given)
∴ PR = RQ 
or AD bisects PQ at R (proved).