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Question 14
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. Show that ΔABCΔPQR.

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Solution


Given, two triangles ΔABC and ΔPQR in which AD and PM are medians such that ABPQ=ACPR=ADPM
To Prove that ΔABCΔPQR
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]
ΔABDΔCED [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]
ΔACEΔPRN [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)]
ΔABCΔPQR [By SAS similarity criterion]





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