Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Prove that `Delta ABC~ Detla PQR`

Explanation

GIVEN AD and PM ar medians of `Delta ABC and Delta PQR` respectively such that
`(AB)/(PQ)=(AC)/(PR)=(AD)/(PM)`
TO PROVE `Delta AB~Delta PQR`
CONSTRUCTION Produce AD to E such that AD= DE and produce PM to N such that PM= MN.
Join EC and NR.

PROOF In `Delta ABD and Delta ECD`, we have
`BD= CD [ :. D` is midpoint of BC]
AD= ED [ by construction]
`angle BDA= angle CDE` [ vertices opposite angels]
`:. Delta ABD~=Delta ECD` [ by SAS- contruency]
And so, `AB= EC" "....(i)` [cpct]
Similarly, `Delta PQM ~= Delta NRM`
and So, `PQ= NR" ".... (ii)` [ cpct]
Now, `(AB)/(PQ)=(AC)/(PR)=(AD)/(PM)` (given)
`rArr (EC)/(NR)=(AC)/(PR)=(AD)/(PM)` [ using (i) and (ii)]
`rArr (EC)/(NR)=(AC)/(PR)=(2AD)/(2PM)=(AE)/(PN) [ :. 2AD =AE, 2 PM= PN]`
`rArr Delta ACE~ Delta PNR` [ SSS- similarity]
`:. angle 2= angle4` [ corresponding angels of similar triangles are equal]
Similarly, `angle 1= angle3`= [ it can be proved by joining BE and QN and showing `Delta ABE~ Delta PQN`]
`:. angle 1 + angle 2 = angle3+angle 4, i.e., angle BAC= angle QPR. " "....(iii)`
Now, in `Delta ABC and Delta PQR`, we have
`(AB)/(PQ)=(AC)/(PR)` [ given]
`angle BACE= angle QPR` [ form (ii)]
`:. Delta ABCD~ Delta PQR`[ by SAS-similarity]