Given, two triangles
ΔABC and
ΔPQR in which AD and PM are medians such that
ABPQ=ACPR=ADPM
To Prove that
ΔABC∼ΔPQR
Construction: Produce AD to E so that AD = DE. Join CE.
Similarly produce PM to N such that PM = MN, also Join RN.
Proof
In
ΔABD and
ΔCDE, we have
AD = DE [By Construction]
BD = DC [
∵ AD is the median]
And,
∠ADB=∠CDE [Vertically opposite angles]
∴ΔABD≅ΔCED [By SAS criterion of congruence]
⇒AB=CE[byCPCT]...(i)
Also, in
ΔPQM and ΔMNR, we have
PM = MN [By Construction]
QM = MR [
∴PM is the median]
And,
∠PMQ=∠NMR [Vertically opposite angles]
∴ΔPQM=ΔMNR [By SAS criterion of congruence]
⇒PQ=RN[CPCT]...(ii)
Now,
ABPQ=ACPR=ADPM
⇒CERN=ACPR=ADPM...[From(i)and(ii)]
⇒CERN=ACPR=2AD2PM
⇒CERN=ACPR=AEPN[∴2AD=AE and 2PM=PN]
∴ΔACE∼ΔPRN [By SSS similarity criterion]
Therefore,
∠2=∠4
Similarly,
∠1=∠3
∴∠1+∠2=∠3+∠4
⇒∠A=∠P...(iii)
Now, in
ΔABCandΔPQR, we have
ABPQ=ACPR(Given)
∠A=∠P[From(iii)]
∴ΔABC∼ΔPQR [By SAS similarity criterion]