If two sides and a median bisecting the third side of a triangle ar respectively proportional to the corresponding sides and median of the other triangle; then the two triangles are similar.

Explanation

Given: `triangleABC and trianglePQR` in which AD and PS are the medians such that.
`(AB)/(PQ)=(AC)/(PR)=(AD)/(PS)`
To prove , `triangleABC ~ trianglePQR`
construciton : produce AD to E such that AD =DE. Join EC. Also produce PS to T such that PS =ST. Join TR.
Proof: In `triangleABD and triangleECD` we have
`{(BD=CD,(because"D is the mid point of BC as AD is the median")),(angle5=angle6,"(vertically opposite angles)"),(AD=DE,"(construction)"):}`
`triangleABD cong triangleECD` (SAS criterion of congruency)
AB=EC (corresponding parts of congruent trinagles) ...(1)
similarly, we can prove
`trianglePQS cong triangleTRS`
PQ = TR .....(2)
Now, since `(AB)/(PQ)=(AC)/(PR)=(AD)/(PS)` (given)
`Rightarrow (EC)/(TR) = (AC)/(PR)=(2AD)/(2PS)` [ from (1) and (2)]
`Rightarrow (EC)/(TR)=(AC)/(PR)=(AE)/(PT)`
`triangleAEC ~ trianglePTR` ( SSS criterion for similarity)
` angle1=angle2` (corresponding angles of similar triangles are equal) ....(3)
Similarly , we can prove
`angle3=angle4`
Adding (3) and (4) , we get
`angle1+angle3=angle2+angle4`
`Rightarrow angleA=angleP`
Now, in `triangleABC and trianglePQR`
`(AB)/(PQ) = (AC)/(PR)`
and `angleA=angleP`
`triangleABC ~ trianglePQR` (SAS criterion for similarity) Hence proved.