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.

We can use a special case of Stewart's Theorem, which has a non-trigonometric proof recently discussed on Stack Exchange.
I will therefore write out a sketch of a trigonometric proof of the result, in the knowledge that the trigonometry can be eliminated.
To make our lives easier, so we don't have to work with ratios, scale one of the triangles to make $AB=PQ$. The other sides that were in the same ratio then also become equal.  So now we want to prove congruence.  Then we can scale back to get similarity.
Look at the left triangle, and let $AB=c$, $AC=B$.  Let $AD=d$, and let $BD=DC=u$.
Consider $\triangle ADB$, and let $\theta$ be the angle at $D$.
By the Cosine Law,
$$c^2=d^2+u^2-2du \cos(\theta).$$
Now in $\triangle ADC$, the angle at $D$ has cosine the negative of $\cos(\theta)$, so again by the Cosine Law,
$$b^2=d^2+u^2+2du\cos(\theta).$$
Add, sneakily letting $\theta$ drop out of the picture. We get 
$$c^2+b^2=2(d^2+u^2).$$
Comment: To get rid of the trigonometry, we can drop a perpendicular to $BC$ from $A$. Then two applications of the Pythagorean Theorem give us the fact that $c^2+b^2=2(d^2+u^2)$, which will turn out to be what we need. (The details of the calculation are done one of the Stewart's Theorem answers referred to earlier.) 
In the other triangle, let $v=QS=SR$.  Remember that the other sides are equal to the corresponding sides of the first triangle, because of the scaling. Thus 
$$c^2+b^2=2(d^2+v^2).$$
It follows that $u=v$, and we are finished.
