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]