Similarity of Triangle problem Given: AD & PS are medians in ΔABC and ΔPQR respectively,
$$\frac{AB}{PQ}=\frac{AD}{PS}=\frac{AC}{PR}$$
To Prove: ΔABC ~ ΔPQR
Figure: 

Problem:
In ΔABD & ΔPQS  or in ΔADC & ΔPSR or in ΔABC & ΔPQR, I have only found that only two sides are proportional but can't figure-out third thing to prove similarity.
Plz help me.

Create a point A' in the direction of AD, such that DA'=AD. Then there is a parallelogram ABA'C. So is parallelogram PQP'R.

And AD/PS=AA'/PP',
So, AB/PQ=AA'/PP'=AC/PR.
=>ΔABA' ~ ΔPQP'
=>∠BAD=∠QPS
So is ∠CAD=∠RPS
Then ∠BAC=∠QPR
=>ΔABC ~ ΔPQR
(Q.E.D.)  
