If two sides and a median bisecting the third side of a Δ are respectively proportional to the corresponding sides and the median of another triangle, then prove that the two triangles are similar. 
                       

Given: ΔABC and ΔPQR where BD and QS are the medians and AB = BC = BD
                                                                                  PQ    QR    QS
To prove: ΔABC ~ ΔPQR
Construction: Produce BD and QS to E and T respectively such that BD = DE and QS = ST. CE and TR  are joined.
Proof: In ΔADB and DCDE,
AD = DC (given)
∠ADB = ∠CDE (Vertically opposite angles)
BD = DE.
∴ ΔADB ≅ ΔCDE (SAS ≅ axiom)
Hence AB = CE and ∠ABD = ∠DEC.
Similarly ΔPQS ≅ Δ RST,
hence PQ = TR and ∠PQS = ∠STR.
Consider Δ EBC and Δ TQR,
BD = 2 BD = BE (from given and construction)-------(1)
QS    2QS    QT
AB = CE and PQ = RT (proved),
      AB = CE ----(2)
      PQ    RT
      AB = BC = BD  (Given)                         -----(3)
      PQ    QR    QS
From (1), (2) and (3),
     BE = CE  = BC
     QT    RT    QR
∴ ΔEBC ~ ΔTQR(SSS similarity axiom).
⇒ ∠DBC = ∠SQR and ∠DEC = ∠STR ---(4) (corresponding angles of similar triangles are proportional)
But ∠ABD = ∠DEC and ∠PQS = ∠STR (proved)-----(5)
∴ ∠ABD = ∠PQS (from (4) and (5)) ----(6)
From (5) and (6),
∠ABC = ∠PQR -----I
In Δ ABC and Δ PQR,
    AB = BC (given)
    PQ    QR
And ∠ABC = ∠PQR (from I)
∴ ΔABC ~ ΔPQR (SAS Similarity).