In triangle PQR,PM is the median , show that PQ+ QR+PR > 2PM
Answers
Answer:
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Step-by-step explanation:
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Answer:
Given that triangle ABC and ΔPQR in which AD and PM are medians drawn on sides BC and QR respectively.
It is given that AB/PQ = AC/PR = AD/PM
We have to prove that ΔABC ~ ΔPQR
Construction: Produce AD to E such that AD = DE and PM to F such that PM = MF
From the figure,
In ΔABD and ΔCDE,
AD = DE [by construction]
∠ADB = ∠CDE [vertically opposite angles]
BD = DC [Since AD is a median]
So, by SAS congruent condition
ΔABC ≅ ΔPQR
AB = CE [by CPCT]
Similarly, we can prove
ΔPQM ≅ ΔRMF
PQ = RF [by CPCT]
Now, given that
AB/PQ = AC/PR = AD/PM
CE/RF = AC/PR = 2AD/2PM
CE/RF = AC/PR = AE/PF [Since AE = AD + DE and AD = DE, Same for PF]
By using SSS Congruent condition
ΔACE ≅ ΔPRF
=> ∠1 = ∠2 ......1
Similarly, ∠3 = ∠4 ......2
Adding equations 1 and 2, we get
∠1 + ∠3 = ∠2 + ∠4
=> ∠A = ∠P
Now, in ΔABC and ΔPQR
AB/PQ = AC/PR
and ∠A = ∠P
By SAS similar condtion,
ΔABC ~ ΔPQR